Two-end solutions to the Allen-Cahn equation in
R3
Changfeng Gui
Department of Mathematics, U-9,
University of Connecticut, Storrs, CT 06269, USA,
e-mail: gui@math.uconn.edu
Yong Liu
School of Mathematics and Physics,
North China Electric Power University, Beijing, China,
e-mail: liuyong@ncepu.edu.cn
Juncheng Wei
Department of Mathematics,
University of British Columbia, Vancouver, BC, V6T 1Z2, Canada,
e-mail: jcwei@math.ubc.ca
February 20, 2015
1
Introduction
The Allen-Cahn equation
−∆u = u − u3 , |u| < 1
(1)
has been studied for several decades and is an important nonlinear PDE due to
the fact that it lies at the interface of several different mathematical fields. The
famous De Giorgi conjecture states that any entire solution to (1) in Rn which
is monotone in one direction should be one dimensional, at least for n ≤ 8.
The conjecture was proved in dimension n = 2 by Ghoussoub-Gui ([13]) and
dimension 3 by Ambrosio-Cabre ([1]), and in dimensions 4 ≤ n ≤ 8 by Savin
([27]), under an additional assumption. For n ≥ 9, counter-examples have been
constructed by del Pino-Kowalczyk-Wei ([11]). Note that monotone solutions
are indeed minimizers with respect to local perturbations ([1]).
A natural extension of De Giorgi’s conjecture is to classify stable or finite
Morse index solutions. Regarding stable solutions, it has been shown ([1], [13])
that stable solutions in R2 are necessarily one-dimensional, while Pacard-Wei
([24]) constructed stable solutions in R8 which are not one-dimensional.
1
The study of finite Morse index solutions is much more involved. In R2 ,
we have now a rather complete picture of Morse index one solutions. They are
so-called four-end solutions, parametrized by the angle between the two lines.
The cross solution, constructed by Dang-Fife-Peletier ([8]), represents a fourend solution with angle π4 , while the almost parallel line solution, constructed
by del Pino-Kowalczyk-Pacard-Wei ([10]), represents a four-end solution with
angle close to π2 or 0. The existence of four-end solutions with any angle between
0 and π2 was proved by two methods: the first through the moduli space theory
by Kowalczyk-Liu-Pacard ([19]), and the second approach by the mountain-pass
variational method by us (Gui-Liu-Wei [15]). It was also shown that the fourend solutions have Morse index one ([15]). On the other hand, monotone and
symmetric properties of a general four end solution have been obtained in [14],
where more general finite morse index solutions in R2 have also been studied
under an extra energy condition or a condition on the asymptotical structure of
nodal curves.
In this paper we are interested in the structure of two-end solutions of (1)
in R3 . It turns out that without the monotone condition, there are actually
a lot of solutions. One simple example is the so called saddle solution whose
nodal sets are precisely the xoy, yoz and xoz planes (Alessio-Montecchiari [3]).
del Pino-Kowalczyk-Wei ([12]) proved that for each non-degenerate minimal
surfaces with finite total curvature, one could find a solution to (1) whose nodal
sets are close to a rescaled version of this minimal surface. In particular, there
are axially solutions whose nodal sets are close to catenoids with very large
waist. Axially symmetric solutions with multiple interfaces which are governed
by the Jocobi-Toda system are constructed in Agudelo-del Pino-Wei [2].
In spite of all these developments, some important questions for (1) in R3
remain unanswered, even for axially symmetric solutions. In this paper, we will
study those axially symmetric solutions which are additionally even with respect
to the xoy plane. In terms of the cylindrical coordinate (r, z) , they satisfy
{
uzz + urr + r−1 ur + u − u3 = 0, r ∈ [0, +∞), z ∈ R,
(2)
u (r, z) = u (r, −z) , ur (0, z) = 0.
Let H (x) = tanh √x2 be the one dimensional heteroclinic solution:
−H ′′ = H − H 3 , H (0) = 0, H (±∞) = ±1.
The solutions we are interested will have H as its asymptotic profile. We say
that a solution u of (2) has growth rate k if it has the following asymptotic
behavior:
∥u (r, ·) − H (· − k ln r + c)∥L∞ (0,+∞) → 0, as r → +∞,
(3)
for certain c ∈ R. The existence results obtained in [2] and [12] based on
Lyapunov-Schmidt reduction arguments
tell
there )are solutions whose
(√ √
) us that
(
growth rate is in the interval
2, 2 + δ and δ −1 , +∞ , where δ is a very
small constant. A natural question is, whether or not there are solutions with
2
[√
]
growth rate in the range
2 + δ, δ −1 . In this paper we answer this question
affirmatively and our main result is the following
(√
)
Theorem 1 For each k ∈
2, +∞ , there exists a solution to (2) which has
growth rate k.
As we will see later, these solutions indeed are monotone in the following
sense:
uz > 0 for z > 0; ur < 0, for r > 0.
(4)
Outside a large ball, the nodal set of the solutions given by Theorem 1 has
two components, each component is asymptotic to a catenoidal end and around
each end, the solution look likes the one dimensional heteroclinic solution. Borrowing a terminology from minimal surface theory, a solution satisfying (2) and
(3) will be called a two-end solution. We emphasize that here by definition the
two-end solutions are axially symmetric. Comparing with the corresponding
definition of two-end minimal surfaces, it seems that a more general definition
of two-end solution should also involve those solutions which are not axially
symmetric and only assume that their nodal set is asymptotic to two catenoidal
ends. In the minimal surface theory, a classical result proved by R. Schoen ([28])
is that a minimal surface with two catenoidal ends is a catenoid. We expect that
the analogous result for Allen-Cahn equation should hold. A major difficulty
we encounter is to show that a solution whose nodal set is asymptotic to two
catenoidal ends is axially symmetric.
Observe that for each k ∈ (0, +∞) , there exists a catenoid with growth rate
k. Taking into account the relation between minimal surface theory and AllenCahn equation, at first glance, one may think that for each k ∈ (0, +∞) , there
should be a two-end solution of Allen-Cahn equation. But this turns out to be
false. In fact, we have
Theorem 2 There does not exist two-end solution with growth rate k ∈ (0,
√
2
2 ].
√ √
Indeed, one expects that for k ∈ ( 22 , 2], two-end solution with growth
)
(√
2, +∞ , there should be a
rate k also should not exist, while for each k ∈
unique
two-end solution with growth rate k. We remark that the lower bound
√
2 is somehow related to the deep facts that the two ends of the solutions to
the Allen-Cahn equation actually “interact” with each other and in this regime
one naturally encounters the so called Toda system, as we will see later in the
analysis. This constitutes a major difference with the theory of minimal surfaces.
Roughly speaking, the Allen-Cahn equation interplays between the theory of
minimal surfaces and the theory of Toda system, which is a classical integrable
system. In the minimal surface theory, the catenoids are basic blocks for the
construction of other minimal surfaces or constant mean curvature surfaces (see
for example [16], [22], [29] and the references therein). It is therefore natural
and interesting to ask what role two-end solutions of the Allen-Cahn equation
play in constructing other solutions. We also remark that the two-end solutions
given by Theorem 1 are all unstable, and the stable solution conjecture says that
3
the only bounded stable solution of the Allen-Cahn equation in R3 should be
one dimensional. It is also expected that the Morse index of two-end solutions
should be equal to one, again, we don’t have a proof of this statement, although
it is known that the two-end solutions constructed in [2] and [12] have Morse
index one.
The results in Theorem 1 could be regarded as a generalization of the corresponding results for four-end solutions in R2 ([18], [19]). To explain this, let
us say a few words about the multiple-end solutions in R2 . By definition, a
2k-end solution of (1) in R2 is a solution whose nodal set outside a large ball is
asymptotic to 2k half straight lines at infinity. It is known (del Pino-KowalczykPacard [9]) that the set of 2k-end solution in R2 has a structure of real analytic
variety of formal dimension 2k. Some examples of solutions near the boundary
of this “moduli space” have been constructed (del Pino-Kowalczyk-Pacard-Wei
[10], Kowalczyk-Liu-Pacard-Wei [20]). As we mentioned above, for k = 2, it
is proved in Kowalczyk-Liu-Pacard [19] that the moduli space of four-end solutions, modulo rigid motion, is diffeomorpic
to the open interval (0, 1) . It is also
)
(
proved there that for each θ ∈ 0, π2 , there exists a four-end solution uθ which
is even with respect to both x and y axis and the asymptotic line of the nodal
set of uθ in the first quadrant makes angle θ with the x axis. Now we observe
that if we reflect the solution to (2) across the z axis, then we get a solution
defined for all (r, z) ∈ R2 and even in both variables. Moreover, the nodal set of
this solution outside a large ball also has four components. Hence the two-end
solutions in R3 are in certain sense analogy of the four-end solutions in R2 . The
equations they satisfied are different from each other only by the term r−1 ur
which makes the problem inhomogeneous. In fact in R2 a major fact used is
that there are two linearly independent kernels ux and uy . The construction of
four-end solution is done by first proving that all four-end solutions are nondegenerate. Here we face the problem of degeneracy. We overcome this difficulty
by applying global bifurcation theory developed by Buffoni, Dancer and Toland
for real analytical variety of formal dimension 1.
Before proceeding to proof of our main results, let us outline the main ideas
of the proof, since this also gives a description of the set of solutions in Theorem
1. The main steps of the proof are similar to the corresponding analysis for fourend solution in dimension two, performed in [18] and [19]. Our basic strategy
is to analyze the structure of the
√ set M of axially symmetric two-end solutions
with growth rate larger than 2. In section 2, we prove that the solutions in
M with some additional natural constraints are compact in certain sense. In
section 3, we show that M has the structure of a real analytic variety of formal
dimension 1. In section 4, we analyze those solutions near the “boundary” of the
moduli space M and prove that they are unique in certain sense. Combining
these results, we conclude the proof by applying a structure theorem for real
analytic varieties. We remark that there are two main differences between the
R2 and R3 case. Firstly, the proof of compactness in 3D case is much more
delicate than the 2D case and detailed asymptotic analysis is needed. Secondly,
as we mentioned, the four-end solutions in 2D are all non-degenerate ([17], [18]),
but we don’t know whether it is true for two-end solutions in 3D.
4
2
Compactness of two-end solutions
The compactness of moduli spaces of minimal surface plays an important role
in the minimal surface theory, for example, it is an important step towards the
classification of certain type of minimal surfaces. We refer to Perez-Ros [26] and
references therein for more details on this and other related subjects. In this
section, we shall investigate the compactness property for two-end solutions of
the Allen-Cahn equation.
Throughout the paper we denote by E the set [0, +∞) × R and by E+ the
set [0, +∞) × [0, +∞). Let {un } be a sequence
of two-end solutions which has
√ √
√
growth rate kn > 2. (Note that for k ∈ ( 22 , 2], we don’t know whether or
not there is a two-end solution with growth rate k.) Then by definition,
∥un (r, ·) − H (· − kn ln r − cn )∥L∞ (0,+∞) → 0, as r → +∞.
(5)
We will show that if the distance of the nodal set of un to the origin is uniformly
bounded with respect to n, then up to a subsequence, {un } converges strongly
to a two-end solution u∞ . Here converging strongly means that un → u∞ in
2
Cloc
(E) and there exist constants k∞ , c∞ such that kn → k∞ , cn → c∞ ,
∥u∞ (r, ·) − H (· − k∞ ln r − c∞ )∥L∞ (0,+∞) → 0, as r → +∞,
and
un − H (z − kn ln r − cn ) → u∞ − H (z − k∞ ln r − c∞ )
in L∞ (E) .
Under the assumption that a solution u has the asymptotic behavior (5) ,
one could actually prove that u has certain monotonicity property.
√
Lemma 3 Suppose u is a two-end solution with growth rate k > 2. Then
∂r u < 0 for r > 0; ∂z u > 0 for z > 0.
(6)
The proof of this result is based on the moving plane method and its proof
will be given in the appendix. By Lemma 3, the two-end solutions we are
analyzing always satisfy (6) . We will denote the nodal set of a solution u in the
upper r-z plane by
{
}
Nu := p ∈ E+ , u (p) = 0 .
Due to the monotonicity property, the set Nu ∩ ∂E+ contains a unique point,
call it Pu .
In the rest of the paper, we use C and α to denote universal constants which
may vary from step to step. The main result of this section is the following
Proposition
√ 4 Let un be a sequence of two-end solutions with growth rate
larger than 2. Assume
|Pun | ≤ C.
Then there exists a two-end solution u∞ such that up to a subsequence {un }
converges strongly to u∞ .
5
The rest of this section is devoted to the proof of Proposition 4.
By Lemma 3, for each solution un , Nun will be the graph of a function
z = fn (r) , r ∈ [tn , +∞).
Here tn satisfies Pun = (tn , fn (tn )) . In particular, if Pun is on the z axis, then
tn = 0.
Lemma 5 Under the assumption of Proposition 4, we have
lim fn (r) = +∞,
r→+∞
uniformly in n.
Proof. We argue by contradiction. If this was not true, there will exist a
constant C0 such that for any l ∈ N, one could find a solution unl satisfying
fnl (r) ≤ C0 , for r ∈ (tnl , l) .
2
Since |un | < 1, the sequence {unl } will converge in Cloc
(E) to a nontrivial
solution W whose nodal set is contained in the strip |z| ≤ C0 . Moreover, by
the monotonicity of un , Wr < 0 for r > 0. The limit w (z) := limr→+∞ W (r, z)
then exists and is a solution of the Allen-Cahn equation in dimension 1 :
−w′′ = w − w3 , in R.
(7)
By the symmetric property of W, w (z) = w (−z) . We also have w (z) → 1, as
z → +∞. But (7) does not have a solution with nodal set containing only in a
bounded set {|z| ≤ C0 }. This is a contradiction and the proof is finished.
By the monotonicity of un , Nun is also the graph of a function over the z
axis:
Nun = {(r, z) : r = gn (z)} .
Similar arguments with slight modification as that of (5) imply that
lim gn (z) → +∞,
z→+∞
also uniformly in n.
To obtain more information on the functions fn , we should use the balancing
formula ([9]). Let X = (0, 0, 1) be the constant vector field on R3 . If u =
u (x, y, z) is a solution to the Allen-Cahn equation, then one could check that
{(
)
}
1
2
div
|∇u| + F (u) X − (∇u · X) ∇u = 0.
2
(
)2
Here F (u) = 41 u2 − 1 . Therefore for each regular domain Ω, we have the
following balancing formula:
)
}
∫ {(
1
2
|∇u| + F (u) X − (∇u · X) ∇u · vds = 0.
(8)
2
∂Ω
Here v is the outward unit normal vector of the boundary ∂Ω.
We would like to use (8) to control the slope of the function fn .
6
Lemma 6 For each ε > 0, there exists tε > 0 such that for all n ∈ N,
fn′ (r2 ) − fn′ (r1 ) < ε for tε < r1 < r2 .
Proof. We first show that for each δ > 0 and r̄ > 0, there exists r∗ > r̄, such
that
fn′ (rn ) < δ, n ∈ N,
for some rn ∈ (r̄, r∗ ). Indeed, if this was not true, then for any r̂ > r̄, there
exists n, such that
fn′ (r) ≥ δ, r ∈ (r̄, r̂) .
(9)
Let r̄ < r1 < r2 < r̂. Consider the region Ω ⊂ R3 given by
{(x, y, z) : z > 0, L1 (r) < z < L2 (r)} ,
where
1
(r − r1 ) ,
(r1 )
1
L2 (r) = fn (r2 ) − ′
(r − r2 ) .
fn (r2 )
L1 (r) = fn (r1 ) −
fn′
By the balancing formula,
)
}
∫ {(
1
2
|∇u| + F (u) X − (∇u · X) ∇u · vds = 0.
2
∂Ω
Using (9) and the fact that u exponentially decays to 1 away from the interface,
we deduce that as r1 → +∞,
{(
)
}
∫
1
2
|∇u| + F (u) X − (∇u · X) ∇u · vds → 0.
2
∂Ω∩{z=0}
On the other hand,
∫
∂Ω∩{z=L1 (r)}
r1 fn′
∼ −c1 √
{(
)
}
1
2
|∇u| + F (u) X − (∇u · X) ∇u · vds
2
(r1 )
,
2
1 + fn′ (r1 )
where c1 is a fixed constant, and
{(
)
}
∫
1
2
|∇u| + F (u) X − (∇u · X) ∇u · vds
2
∂Ω∩{z=L2 (r)}
r2 fn′ (r2 )
∼ c1 √
.
2
1 + fn′ (r2 )
7
Combining these estimates, we get a contradiction if r2 is large enough.
Fix an ε > 0, if the conclusion of the lemma was not true, then for any l > 0,
one could find l < r1 < r2 such that
fn′ (r2 ) − fn′ (r1 ) = ε
(10)
and
1
3
ε < |fn′ (r)| < ε, r ∈ (r1 , r2 ) .
2
2
In this case, similarly as above, we could estimate
)
}
{(
∫
1
2
|∇u| + F (u) X − (∇u · X) ∇u · vds
2
∂Ω∩{z=0}
= o (1) r1 .
Therefore the balancing formula tells us that
fn′ (r2 ) r2 − fn′ (r1 ) r1 = o (1) r1 .
This contradicts with (10) .
Lemma 6 tells us that the slope of the nodal line could not increase much as
r increases. This in particular implies the following
Lemma 7 Under the assumption of Proposition 4, we have
lim fn′ (r) = 0,
r→+∞
uniformly in n.
Proof. We argue by contradiction and assume that there exists δ > 0 such that
for each C0 > 0, one could find t0 > C0 and n satisfying
fn′ (t0 ) > δ.
Then one could find t̄1 , t̄2 , such that
(
)
δ
′
fn (r) ∈
, δ , r ∈ (t̄1 , t̄2 ) ,
2
δ
fn′ (t̄1 ) = , fn′ (t̄2 ) = δ.
2
This contradicts with Lemma 6.
Since |un | < 1, one could assume without loss of generality that un → u∞
2
in Cloc
(E) for a solution u∞ . Note that u∞ could not be identically 0. By the
monotonicity property of un , u∞ is also monotone, that is, u∞ satisfies (4). We
need to prove un → u∞ in the strong sense, which in particular implies that u∞
is a solution to (2) whose nodal line is asymptotic to a log curve at infinity and
therefore a two-end solution.
8
The previous lemmas yield certain information on the nodal curve Nun , but
we still don’t have precise description of their shape. On the other hand, we
know that in the region where r is large, locally around the nodal line, the
solution resembles the heteroclinic solution. Our main idea to prove the strong
convergence of {un } is to get a precise decay estimate of the solution un to H
along their nodal lines. For notational simplicity, sometimes we will drop the
subscript n for the function un , fn .
To proceed, let us introduce the Fermi coordinate around the smooth curve
z = f (·) . This coordinate, denoted by (r1 , z1 ) , is defined through the relation
(r, z) = (r1 , f (r1 )) + z1 n,
where n is the unit normal vector at the point (r1 , f (r1 )) of the curve z = f (·) ,
upward pointed. Explicitly,

′
 r = r1 − z1 √ f
,
1+f ′2
(11)
1
 z = f + z1 √ ′2 .
1+f
Here the function f is evaluated at the point r1 . The map X : (r1 , z1 ) →
(r, z) from the Fermi coordinate to the original coordinate system will be a
diffeomorphism in certain tubular neighborhood of the graph of f.
From (11), we get
(
) (
)−1
∂r r1 ∂z r1
∂r1 r ∂z1 r
=
∂r z1 ∂z z1
∂r1 z ∂z1 z

−1
′
B − √ f ′2
1+f

= ′
√ 1 ′2
fB
1+f


′
=
Here
B =1−
f
(1+f ′2 )B
√ 1 ′2
1+f
1
(1+f ′2 )B
′
− √ f ′2
1+f
z1 f ′′
3
(1 + f ′2 ) 2
.
(12)
.
Let us denote by ∆(r,z) = ∂r2 + ∂z2 and ∆(r1 ,z1 ) = ∂r21 + ∂z21 . These two operators
are related by
(
)
1
∆(r,z) = ∆(r1 ,z1 ) +
− 1 ∂r21
A
1 ∂z 1 A
1 ∂r 1 A
+
∂z 1 −
∂r 1 ,
(13)
2 A
2 A2
where
f ′′2
f ′′
+ z12
A = 1 + f ′2 − 2z1 √
2.
(1 + f ′2 )
1 + f ′2
9
(14)
This follows from direct computation, see [10] for more details. From formula
(13), one sees that if f ′ , f ′′ , f (3) is small, then ∆(r,z) is a small perturbation
of ∆(r1 ,z1 ) . One should keep in mind that in (13) , there are terms (in ∂r1 A)
involving the third derivatives of f and these terms should be dealt with very
carefully.
For any function Θ (r, z), the pullback of Θ by the diffeomorphism X will
be defined as
X ∗ Θ (r1 , z1 ) := Θ ◦ X (r1 , z1 ) .
Occasionally, X ∗ Θ will also be denoted by Θ∗ . Keep in mind that X ∗ Θ is only
defined in the region where the (Fermi) coordinate is well defined. Conversely, for
∗
any function Θ (r1 , z1 ), we set X −1 Θ (r, z) = Θ ◦ X −1 (r, z) .
To prove the compactness, we need to know the precise asymptotic behavior
of un . We shall define suitable approximate solutions and analyze the error
between the approximate and true solutions. It turns out that to get a good
approximate solution, it will be convenient to use the Fermi coordinate. Clearly
there is a technical issue concerning the Fermi coordinate. Namely, the Fermi
coordiante is in general not well defined in the whole r-z plane. Note that
for r1 large, by Lemma 7, f ′ is small. We also know that u is close to the
one dimensional heteroclinic solution locally around the nodal line at far away.
Hence intuitively, for r large, f ′′ ,f (3) , ..., should also be small. Indeed, we have
Lemma 8 As r tends to infinity, |f ′′ (r)| , f (3) (r) , f (4) (r) tend to zero.
Proof. Let r0 be a fixed large constant. Around the point (r0 , f (r0 )) , let us
consider the line l1
z − f (r0 ) = f ′ (r0 ) (r − r0 ) .
Let l2 be the line orthogonal to it and passing through (r0 , f (r0 )) . Use these
two lines to build an orthogonal coordinate system, denoted by (s, t) , where
the s coordinate corresponding to line l1 . We know that u is close to H (t) . By
analyzing the equation satisfied by the error ϕ (r, z) := u (r, z) − H (t) , we find
that for r0 large, around (r0 , f (r0 )) , the C 2 norm of ϕ is small.
Observe that u (r, f (r)) = 0. Hence using the definition of ϕ,


′
f
(r)
−
[f
(r
)
+
f
(r
)
(r
−
r
)]
0
0 
√ 0
H
+ ϕ (r, f (r)) = 0.
(15)
2
′
1 + (f (r0 ))
Differentiate this equation with respect to r, using the fact that ∂z ϕ is small, we
get |f ′ (r) − f ′ (r0 )| is small, which we already know. Differentiate (15) twice,
we obtain
H ′ (·) f ′′ (r) + H ′′ (·) (f ′ (r) − f ′ (r0 )) + ∂r2 ϕ + 2∂r ∂z ϕf ′ + ∂z ϕf ′′ = 0.
2
Using the fact that ∂z ϕ is small, we deduce f ′′ (r0 ) is small.
Indeed, one could continue this process and show that the higher order
derivatives f (3) , f (4) are also small.
10
Since |f ′′ | is small, the Fermi coordinate actually will be well defined in
a very large tubular neighborhood of Nu . For later purposes, we need to be
more precise in describing the size of the region where Fermi coordinate is well
defined.
For each point Z = (r, z) ∈ E, we use dist (Z, Nu ) to denote the distance
between Z and the curve Nu . That is,
dist (Z, Nu ) = min {|Z − p| , p ∈ Nu } .
Let π (Z) be the set of points which realize this distance. Hence
π (Z) = {p ∈ Nu : |Z − p| = dist (Z, Nu )} .
For each r1 , let us consider the set
Dr1 := {Z : π (Z) = {(r1 , f (r1 ))}} .
Note that by the triangle inequality, Dr1 is connected. Then
Dr1 = {X (r1 , t) : t ∈ (−d1 (r1 ) , d2 (r1 ))} ,
for some functions d1 and d2 (di may not be differentiable).
Define
d¯(r1 ) = min {d1 (r1 ) , d2 (r1 ) , 3f (r1 )}
and
{
(
)}
D̄r1 = X (r1 , t) : t ∈ −d¯(r1 ) , d¯(r1 ) .
Here the constant 3 has not particular importance. Fix a large constant r0 . Let
Bu := ∪r1 >r0 D̄r1 . It is worth to be pointed out that in principle the boundary of
Bu could be quite complicated. Let η, whose existence is related to assumption
(A) below, be a smooth cutoff function equals 0 outside ∪r1 >r0 −1 D̄r1 and
(
)
X ∗ η (r1 , z1 ) = 1, for z1 ∈ −d¯(r1 ) + 1, d¯(r1 ) − 1 , r1 > r0 .
Similarly, let Bu+ = Bu ∩E+ and we define a similar cutoff function η + supported
in Bu+ (note that the notation η + does not mean the positive part of η).
Since Bu is not the whole r-z plane, we shall use the cutoff function η to
define a smooth function H1 by the formula
H1 = ηH1 + (1 − η)
H1
.
|H1 |
Here the function H1 is defined through
X ∗ H1 (r1 , z1 ) = H (r1 − h (z1 )) ,
where h is a small function to be determined later. We emphasize that since we
are only interested in the behavior of u outside of a bounded set, H1 should be
regarded as a function only defined on the set Ê, where
}
{
1
1
(r − r0 ) < z < f (r0 ) − ′
(r − r0 ) .
Ê = E\ (r, z) : −f (r0 ) + ′
f (r0 )
f (r0 )
11
For any function Θ, let Θs (r, z) = Θ (r, −z) .
With all the previous notations introduced, we now define an approximate
solution ū as
ū = H1 + H1s + 1.
The function ū is defined on the set Ê, rather than the whole plane E. Let
ϕ = u − ū be the difference between the true solution u and the approximate
solution ū. Note that since ū is even with respect to the z variable, ϕ is also
even. Introduce the function H1′ by
X ∗ H1′ (r1 , z1 ) = H ′ (z1 − h (r1 )) .
Then the small function h appeared in the definition of the function H1 is
required to satisfy the following orthogonality condition:
∫
(
)
X ∗ ϕη + H1′ dz1 = 0, r1 > r0 .
(16)
R
The existence of h is guaranteed by the following:
Lemma 9 There exists a small function h(in C 2 sense) satisfying (16) .
Proof. This follows from similar arguments as that of [19]. The basic idea is to
use the fact that ϕ is small and apply the implicit function theorem. We omit
the details.
One advantage of using the Fermi coordinate with respect to the nodal curve
is that h could be estimate in terms of ϕ and f. To see this, let N s (u) be the
graph of the function z = −f (r) . For Z = (r, z) ∈ E, let
D (r, z) = dist (Z, N (u)) + dist (Z, N s (u)) ,
that is, the sum of the distance of Z to the nodal line in the upper plane and
lower plane. Set D (r1 ) = D (r1 , f (r1 )) .
Lemma 10 For r1 > r0 ,
|h (r1 )| ≤ C |X ∗ ϕ (r1 , 0)| + Ce−
√
2D(r1 )
.
Proof. Since ϕ = u − ū, around Nu we have
X ∗ ϕ (r1 , z1 ) = X ∗ u (r1 , z1 ) − [H (z1 − h (r1 )) + X ∗ H1s + 1] .
(17)
Setting z1 = 0 in the above equation, using the fact that X ∗ u (r1 , 0) = 0, we
find
X ∗ ϕ (r1 , 0) + H (−h (r1 )) + 1 + X ∗ H1s (r1 , 0) = 0.
On the other hand, by the definition of H1s and the asymptotic behavior of H,
|1 + X ∗ H1s (r1 , 0)| ≤ Ce−
12
√
2D(r1 )
.
Therefore,
|h (r1 )| ≤ C |X ∗ ϕ (r1 , 0)| + Ce−
√
2D(r1 )
.
This completes the proof.
Following the above arguments and differentiate equation (17) with respect
to r1 and setting z1 = 0 in the obtained equation, we get
|h′ (r1 )| ≤ C |∂r1 X ∗ ϕ (r1 , 0)| + Ce−
√
2D(r1 )
,
Similar arguments yield estimates for the higher order derivatives and Holder
norms.
One of the main ingredients in the proof of Proposition 4 is to get suitable
decay estimate for the function ϕ. In this respect, we shall first of all prove the
following estimate.
Proposition 11 For r1 > r0 , the function ϕ satisfies
∥X ∗ ϕ (r1 , ·)∥L∞ ≤ Ce−r1 + Ce−D(r1 ) +
C
.
r12
We shall first of all prove this proposition under an additional assumption
on the size of the Fermi coordinate. More precisely, at this stage, we assume
that
′ 1 ′′ d¯ ≤ , d¯ ≤ C and d¯(r) ≥ 2D (r) .
(A)
2
3
Later we will indicate the necessary modification of the proof if this assumption
is not a priori satisfied.
Clearly ϕ satisfies
(
)
Lϕ := −∆ϕ + 3ū2 − 1 ϕ = E (ū) + P (ϕ) .
(18)
Here the notation E (ū) stands for
ūrr + r−1 ūr + ūzz + ū − ū3 ,
which is the error of the approximate solution ū, and
P (ϕ) = −3ūϕ2 − ϕ3
is a higher order perturbation term.
For any function Θ, we define the projection of Θ onto η + H1′ as
∫
X ∗ (Θη + H1′ ) dz1 + ′
∥
)
Θ1 = ∫ R (
η H1 .
∗ (η + )2 H ′2 dz
X
1
1
R
Since the solutions are even in the z variable, we also define
[ ]s
∥
∥
Θ∥ = Θ1 + Θ1 .
13
Finally, set
Θ⊥ := Θ − Θ∥ .
Equation (18) could be written in the form
⊥
∥
Lϕ = [E (ū)] + [E (ū)] + P (ϕ) .
∥
As we will see later, [E (ū)] is small(in suitable norm) compared to ϕ, and ϕ is
⊥
then essentially controlled by [E (ū)] . As a crucial step towards the proof of
Proposition 11, we shall take the task of analyzing E (ū) . By definition
ū = H1 + H1s + 1.
A simple manipulation leads to the following expansion
(
)
2
E (ū) = E (H1 ) + E (H1s ) + 3 (H1s + 1) H12 − 1 + 3 (H1 + 1) (H1s + 1) . (19)
Certainly the function E (ū) is also even in the z variable. One expects that in
the upper plane E+ , the main
( order) of E (ū) should be E (H1 ) and one of the
interaction term 3 (H1s + 1) H12 − 1 .
We first analyze the projection of E (H1 ) onto η + H1′ . For technical reasons,
we introduce a small perturbation of f √
due to the presence of the modulation
function ∫h. Let p = f + ĥ, where ĥ := 1 + f ′2 h. Throughout the paper, we
set c0 = R H ′2 (t) dt. The notation O (g) represents a function in r1 such that
|O (g)| ≤ C |g (r1 )| .
Lemma 12 For r1 > r0 ,
∫
(
)′
( + ′
)
( )
c0
r1 p′
√
X η H1 E (H1 ) dz1 = (1 + o (1))
+ O (h′′ f ′′ ) + O h′2
′2
r1
1+p
R
( √
)
(
)
¯
+O (h′ f ′′ ) + O h′ r−1 + O e−2 2d(r1 ) + O (h′ h′′ ) + O (hf ′′ )
( ′3 )
(
)
f
.
+ O (hf ′′′ ) + O f ′′3 + O
r13
∗
Proof. In the region where the cutoff function η + ̸= 0, E (H1 ) could be expressed in terms of the Fermi coordinate (r1 , z1 ) :
E (H1 ) = ∆(r,z) H1 + r−1 ∂r H1 + H1 − H13
= A−1 ∂r21 H1 + ∂z21 H1
∂z A
∂r A
+ 1 ∂z1 H1 − 1 2 ∂r1 H1
2A
2A
+ r−1 (∂r1 H1 ∂r r1 + ∂z1 H1 ∂r z1 ) + H1 − H13 .
Obviously ∂z1 H1 = H ′ , ∂z21 H1 = H ′′ , and
∂r1 H1 = −h′ H ′ , ∂r21 H1 = −h′′ H ′ + h′2 H ′′ ,
14
where H ′ and H ′′ are evaluated at z1 − h (r1 ) . Since H ′′ + H − H 3 = 0, we
obtain
(
)
( ′′ ′
)
∂z1 A
−1
′2 ′′
−1
E (H1 ) = A
−h H + h H +
+ r ∂r z1 H ′
2A
(
)
∂r 1 A
−1
+ −
+ r ∂r r1 (−h′ H ′ ) .
(20)
2A2
It follows that
∫
(
)
X ∗ η + H1′ E (H1 ) dz1
R
(
)
∫
∫
(
)
∂z 1 A
=
η +∗ A−1 −h′′ H ′ + h′2 H ′′ H ′ + η +∗
+ r−1 ∂r z1 H ′2
2A
R
R
(
)
∫
(
)
√
∂r 1 A
¯
+ η +∗
− r−1 ∂r r1 h′ H ′2 + O e−2 2d
2
2A
R
(
)
∫
∫
∂z 1 A
= −h′′ η +∗ A−1 H ′2 + η +∗
+ r−1 ∂r z1 H ′2
2A
R
R
( √ )
( ′2 )
(
)
¯
′ ′′
′ ′′′
+ O h + O (h f ) + O (h f ) + O h′ r−1 + O e−2 2d .
( √ )
¯
The appearing of the term O e−2 2d is due to the fact that in the upper
( √ )
boundary of Bu , E (H1 ) = O e− 2d .
To proceed, we shall calculate the second term in the above expression, which
roughly speaking should be the main order term of the projection. We have
∂z 1 A
+ r−1 ∂r z1
2A
′′
f ′′2
− √ f ′2 + z1 (1+f
′2 )2
f′
1+f
)
−(
=
√
A
′
1 + f ′2
r1 − z1 √ f ′2
1+f
(
)(
)
( ′′2 )
f ′′
f ′′2
f ′′
= −
1 + 2z1
3 + z1
3 + O f
3
(1 + f ′2 )
(1 + f ′2 ) 2
(1 + f ′2 ) 2
(
( ′2 ))
f′
f′
f
− √
1 + z1 √
+O
′2
′2
r12
r1 1 + f
r1 1 + f
(
)
f ′′
z1 f ′′2
=−
+ O f ′′3
3 −
3
(1 + f ′2 )
(1 + f ′2 ) 2
( ′3 )
f′
z1 f ′2
f
− √
− 2
+
O
.
r1 (1 + f ′2 )
r13
r1 1 + f ′2
15
Using this estimate, we get
∫
(
)
X ∗ η + H1′ E (H1 )
R
]
[
c0 h′′
c0 f ′′
c0 f ′
√
+
= − (1 + o (1))
3 +
1 + f ′2
r1 1 + f ′2
(1 + f ′2 ) 2
)
( )
(
+ O h′2 + O (h′ f ′′ ) + O (h′ f ′′′ ) + O h′ r−1
( √
)
¯
+ O (h′′ f ′′ ) + O e−2 2d(r1 ) .
(21)
In this expression, we are mainly interested in the first term. We calculate
r1 h′′
r1 f ′′
f′
+
3 + √
′2
1+f
1 + f ′2
(1 + f ′2 ) 2
[
(
)′
(
)′′ ]
r1
ĥ′′
1
1
′
√
=
+ 2ĥ √
+ ĥ √
1 + f ′2
1 + f ′2
1 + f ′2
1 + f ′2
+
r1 f ′′
3
2
+√
f′
1 + f ′2
(1 + f ′2 )
(
)′
r1 p′
= √
+ O (h′ r1 h′′ ) + O (h′ r1 f ′′ )
1 + p′2
+ O (hr1 f ′′ ) + O (hr1 f ′′′ ) + O (h′ ) .
The conclusion of the lemma then follows from this estimate.
With the projection being understood, we proceed to analyze the orthogonal
part.
Lemma 13 For r1 > r0 , the following estimate is true:
)
(
f ′′2
f ′2
∥
H′
E (H1 ) − [E (H1 )]1 = − (z1 − h)
3 + r 2 (1 + f ′2 )
(1 + f ′2 )
1
( √ )
(
)
( )
′′ ′′
− 2D
+ O (h f ) + O e
+ O h′ r−1 + O h′′2
( ′3 )
( ′′3 )
( )
f
+O f
+O
+ O h′2 + O (h′ f ′′ ) + O (h′ f ′′′ ) .
r13
(22)
16
Proof. Let us consider typical terms appeared in (20) .
[
]∥
A−1 h′′ H ′ − A−1 h′′ H ′ 1
∫ −1 ′′ +∗ ′2
A h η H +∗ ′
−1 ′′ ′
= A h H − R∫
η H
2
(η +∗ H ′ )
R
[
]
∫
h′′ H ′
η +∗
+∗ ′2
=
1− ∫
η H
+ O (f ′′ h′′ )
+∗ H ′ )2
1 + f ′2
(η
R
R
)
(∫
∫
(
)
2
η +∗ H ′ − η +∗ η +∗ H ′2 H ′ + O (f ′′ h′′ ) .
= O (h′′ )
R
Note that
(∫
(
R
Hence
η
+∗
H
R
)
′ 2
∫
−η
+∗
η
+∗
H
′2
)
( √ )
H ′ = O e− 2f .
R
( √ )
[
]∥
( )
A−1 h′′ H ′ − A−1 h′′ H ′ 1 = O h′′2 + O e− 2D + O (f ′′ h′′ ) .
Next, we have calculated that
∂z 1 A
f ′′
z1 f ′′2
+ r−1 ∂r z1 = −
3 −
3
2A
(1 + f ′2 )
(1 + f ′2 ) 2
f′
z1 f ′2
√
− 2
r1 (1 + f ′2 )
r1 1 + f ′2
( ′3 )
(
)
f
+ O f ′′3 + O
.
r13
−
Using this expansion, we find
(
)
[(
) ]∥
∂z 1 A
∂z 1 A
+ r−1 ∂r z1 H ′ −
+ r−1 ∂r z1 H ′
2A
2A
1
(
)
′′2
′2
f
f
H′
= − (z1 − h)
3 + r 2 (1 + f ′2 )
(1 + f ′2 )
1
( ′3 )
( √ )
( ′′3 )
f
+O f
+O
+ O e− 2D .
r13
The rest of the terms always contains small order terms of h′ or h and the
conclusion of the lemma readily follows.
The next result gives us an estimate for the main interaction term between
H1 and H1s appeared in E (ū) . Let
√
√ ∫
c1 = 3 2 H ′2 (s) e 2s ds.
R
17
Lemma 14 For r1 > r0 ,
∫
√
[
(
)]
X ∗ 3η + H1′ (H1s + 1) H12 − 1 dz1 = −c1 (1 + o (1)) e− 2D(r1 ) .
R
√
Proof. Since H ′′ = H 3 − H, H12 − 1 = − 2H1′ , it follows that
∫
[
(
)]
X ∗ 3η + H1′ (H1s + 1) H12 − 1 dz1
R
( √ )
√ ∫
[
]
= −3 2 X ∗ η + H1′2 (H1s + 1) dz1 + o e− 2D
R
√ ∫
(
) √
= −3 2 X ∗ η + H1′2 e− 2|z2 | dz1
(∫ R
)
( + ′2 ) −2√2|z2 |
∗
X η H1 e
+O
dz1 .
R
Since by definition |z2 | = D (r, z) − |z1 | , we find
∫
[
(
)]
X ∗ 3η + H1′ (H1s + 1) H12 − 1 dz1
R
√ ∫
(
) √ ∗
= −3 2 X ∗ η + H1′2 e− 2(X D−|z1 |) dz1
(∫ R
)
√
(
)
∗
+O
X ∗ η + H1′2 e−2 2(X D−|z1 |) dz1
R
√ ∫
(
) √ ∗
= −3 2 X ∗ η + H1′2 e− 2(X D−|z1 |) dz1
)
( √ R
+ o e− 2D(r1 )
∫
√
√
(
) √
= −3 2e− 2D(r1 ) X ∗ H1′2 e 2|z1 | dz1
R
( √
)
− 2D(r1 )
+o e
.
In the last equality, we have used the fact that in E+ ,
|X ∗ D (r1 , z1 ) − D (r1 )| = o (z1 ) .
Remark 15 Clearly the function D (r1 ) is strictly less than 2f (r1 ) . However,
due to the fact that f ′ is small,
2f (r1 ) − D (r1 ) = o (f (r1 )) .
It is exactly this fact that makes it possible for us to analyze the equation satisfied
by f.
18
The previous result analyze directly the error E (ū) from the definition of ū.
In the next result, we will express the projection of E (ū) onto η + H1′ in terms
of the function ϕ. The main equation we will use is equation (18) satisfied by ϕ.
Lemma 16 For each r1 > r0 ,
∫
(
)
X ∗ η + H1′ E (ū) dz1 R
(
)
)
(
√
2
= O ∥ϕ∗ (r1 , ·)∥∞ + O ∥ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O (∥ϕ∗ (r1 , ·)∥∞ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
(
)
(
)
√
)
(
2
+O ∥∂r1 ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O ∥ϕ∗ (r1 , ·)∥∞ ∂r21 ϕ∗ (r1 , ·)∞
+O (f ′′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ ) + O (f ′′ ∥∂r1 ϕ∗ (r1 , ·)∥∞ ) + O (f ′′′ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
(
)
(
)
+O r−1 ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O r−1 f ′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ .
Proof. Multiplying both sides of (18) with η + H1′ and integrating in R, we get
∫
∫
∫
(
)
(
)
(
)
X ∗ η + H1′ E (ū) dz1 =
X ∗ η + H1′ Lϕ dz1 − X ∗ η + H1′ P (ϕ) dz1 .
R
∫
R
R
∗
+
H1′ Lϕ) dz1 .
By formula (13) of Laplacian
Let us calculate the term R X (η
in the Fermi coordinate,
∫
(
)
X ∗ η + H1′ Lϕ dz1
R
[
]
∫
(
)
1 ∂z1 A
1 ∂r 1 A
= − X ∗ η + H1′ A−1 ∂r21 +
∂z1 −
∂
ϕ∗ dz1
r
1
2 A
2 A2
R
∫
(
)
− X ∗ η + H1′ r−1 [∂r1 ϕ∗ ∂r r1 + ∂z1 ϕ∗ ∂r z1 ] dz1
∫R
[
(
)
(
(
) )]
+
−X ∗ η + H1′ ∂z21 ϕ∗ + X ∗ η + H1′ 3ū2 − 1 ϕ dz1 .
R
We first estimate
∫
R
X ∗ (η + H1′ ) A−1 ∂r21 ϕ∗ dz1 . Differentiating the identity
∫
(
)
X ∗ η + H1′ ϕ dz1 = 0
(23)
R
with respect to r1 , we obtain
∫
∫
(
)
[
(
)]
X ∗ η + H1′ ∂r1 ϕ∗ dz1 = − ∂r1 X ∗ η + H1′ ϕ∗ dz1
R
∫R
∫
( +∗ ) ′ ∗
= − ∂r 1 η
H ϕ dz1 +
η +∗ H ′′ h′ ϕ∗ dz1 .
R
′
R
′′
Here H and H are evaluated at z1 − h (r1 ) . Therefore we could apply Lemma
10 and obtain
∫
(
)
X ∗ η + H1′ ∂r1 ϕ∗
R
)
(
√
= O ∥ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O (∥ϕ∗ (r1 , ·)∥∞ ∥∂r1 ϕ∗ (r1 , ·)∥∞ ) .
19
Similarly, differentiating (23) with respect to r1 twice, and using the fact that
1
′′
A−1 − 1+f
′2 = O (f ) , we get
∫
(
)
X ∗ η + H1′ A−1 ∂r21 ϕ∗ dz1
R
∫
(
)
(
)
1
=
X ∗ η + H1′ ∂r21 ϕ∗ dz1 + O ∂r21 ϕ∗ (r1 , ·)∞ |f ′′ (r1 )|
′2
1+f
R
∫
∫
[ ∗ ( + ′ )]
[
(
)]
1
2
∗
∂r1 X η H1 ∂r1 ϕ −
∂r21 X ∗ η + H1′ ϕ∗
=−
′2
′2
1+f
1+f
R
R
(
)
+ O ∂r21 ϕ∗ (r1 , ·)∞ |f ′′ (r1 )|
)
(
(
)
√
2
= O ∥∂r1 ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O ∥∂r1 ϕ∗ (r1 , ·)∥∞
(
)
√
+ O ∥ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O (∥ϕ∗ (r1 , ·)∥∞ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
)
(
+ O ∥ϕ∗ (r1 , ·)∥∞ ∂r21 ϕ∗ (r1 , ·)∞ .
Next, using ∂z1 A = O (f ′′ ) , we infer
∫
(
)
∂z1 A
∂z1 ϕ∗ X ∗ η + H1′ dz1 = O (f ′′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ ) .
A
R
Observe that ∂r1 A = O (|f ′′ | + |f ′′′ |) , thus
∫
(
)
∂r 1 A
∂r1 ϕ∗ X ∗ η + H1′ dz1 = O (|f ′′ | ∥∂r1 ϕ∗ (r1 , ·)∥∞ )+O (|f ′′′ | ∥∂r1 ϕ∗ (r1 , ·)∥∞ ) .
2
R A
We also have
∫
(
)
r−1 X ∗ η + H1′ [∂r1 ϕ∗ ∂r r1 + ∂z1 ϕ∗ ∂r z1 ]
R
(
)
(
)
= O r−1 ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O r−1 |f ′ | ∥∂z1 ϕ∗ (r1 , ·)∥∞ .
Another term we need to estimate is
∫
∫
(
)
(
(
) )
X ∗ η + H1′ ∂z21 ϕ∗ +
X ∗ η + H1′ 3ū2 − 1 ϕ
R
R
∫
∫
) )
(
(
)
(
2 ∗
∗
+ ′
=
X η H1 ∂z1 ϕ + X ∗ η + H1′ 3H12 − 1 ϕ
R
R
∫
( + ′( 2
) )
∗
2
+ 3 X η H1 ū − H1 ϕ .
R
To handle them, we integrate by parts for the first term, and for the last term
we use the fact
(
)
(
)
(
)
η + ū2 − H12 H1′ = O η + (H1s + 1) H1′ = O e−D .
We conclude that
∫
∫
(
)
(
(
) )
X ∗ η + H1′ ∂z21 ϕ∗ + X ∗ η + H1′ 3ū2 − 1 ϕ
R
R
(
√ )
∗
− 2f
+ O (∥ϕ∗ (r1 , ·)∥∞ ∥∂r1 ϕ∗ (r1 , ·)∥∞ ) .
= O ∥ϕ (r1 , ·)∥∞ e
20
Finally, since P (ϕ) is higher order term of ϕ, the estimate of the term
∗
X
(η + H1′ P (ϕ)) dz1 is trivial. Combining all these estimates, we get the
R
desired result.
With the error term E (ū) being understood, we would like to recall some
properties of the operator L. The next result are essentially proved in [10],
although here we need to do slight modifications due to the presence of boundary
terms. For each r1 ≥ r0 , let
}
{
1
1
(r − r1 ) < z < f (r1 ) − ′
(r − r1 ) ,
Kr1 := (r, z) ∈ E, −f (r1 ) + ′
f (r1 )
f (r1 )
∫
and Jr1 := E\Kr1 . Note that Jr0 is simply the set Ê introduced before.
Lemma 17 Suppose φ is a solution of the equation
Lφ = Θ in Jr1 .
∥
and φ (r, z) = φ (r, −z) , φ1 = 0. Then
|φ∗ (s, z)| ≤ C ∥φ∥L∞ (∂Jr ) er1 −s + ∥Θ∥L∞ (Jr ) , for each s > r1 .
1
1
Proof. Let θ be a cutoff function such that
{
1, if Z ∈ Jr1 and dist (Z, ∂Jr1 ) > 1,
θ (Z) =
0, Z ∈
/ Jr1 .
Consider the function φ̄ := θφ. Then φ̄ satisfies
Lφ̄ = θΘ + ∆θφ + 2∇θ∇φ.
(24)
Modify the function ū in E\Jr1 if necessary, we could get a function, still denoted
by ū, whose nodal curve are almost horizontal and around its nodal curve,
it looks like the heteroclinic function H. With slight abuse of notation, the
corresponding linearized operator will be still write as L. We could also assume
that φ̄ satisfies the orthogonality condition.
Denote Θ1 = θΘ, Θ2 = ∆θφ + 2∇θ∇φ. For i = 1, 2, consider the equation
(
)s
Lφ̄i = Θi + ki (r1 ) η + H1′ + ki η + H1′
(25)
∫
where φ̄i , k̄i are unknown functions and we require( R X)∗ (η + Hi′ φ̄i ) dz1 = 0, i =
1, 2. By the results of [10], one could find solution φ̄i , k̄i to this equation. This
pair of solution is indeed unique and hence we have the following estimate for
φ̄1 :
∥φ̄1 ∥L∞ ≤ C ∥Θ1 ∥L∞ .
For the function φ̄2 , since Θ2 has better decay property than Θ1 , we could
estimate
|φ̄2 (r, z)| ≤ C ∥Θ2 ∥L∞ er1 −r .
21
Notice that
[
]s
L (φ̄1 + φ̄2 ) = Θ1 + Θ2 + (k1 + k2 ) η + H1′ + (k1 + k2 ) η + H1′ .
But the equation
[
]s
Lφ̄3 = Θ1 + Θ2 + k3 η + H1′ + k3 η + H1′
∫
also has a unique pair of solution (φ̄3 , k3 ) if we require R X ∗ (η + H1′ φ̄3 ) = 0.
This implies that φ̄ = φ̄1 + φ̄2 , the conclusion of this proposition readily follows.
Once we have an estimate in L∞ norm, we could get a corresponding Holder
norm estimate by Schauder’s elliptic estimate. Now we proceed to the proof of
Proposition 11.
Proof of Proposition 11. Recall that
∫
(
)
X ∗ η + H1′ E (ū)
R
∫
∫
(
)
(
(
))
=
X ∗ η + H1′ E (H1 ) + X ∗ η + H1′ 3 (H1s + 1) H12 − 1
∫R
∫R
(
)
(
)
2
+ X ∗ η + H1′ E (H1s ) + X ∗ η + H1′ 3 (H1 + 1) (H1s + 1) .
R
R
We have already analyzed the first two terms. For the third term, slightly abuse
the notation, we have
∫
∫
(
)
(
)
(
)
X ∗ η + H1′ E (H1s ) dz1 =
X ∗ η + H1′ A−1 −h′′ H ′ + h′2 H ′′ dz1
R
(
)
∫R
( + ′ ) ∂z 2 A
∗
−1
+
X η H1
+ r ∂r z2 H ′ dz1
2A
R
(
)
∫
( + ′ ) ∂r 2 A
∗
−1
+
X η H1
− r ∂r r2 (h′ H ′ ) dz1 .
2A2
R
It should be pointed out that here A, h are evaluated at r2 and H ′ is evaluated
at z2 − h (r2 ) , and h′ means the derivative with respect to r2 ((r2 , z2 ) being the
Fermi coordinate with respect to z = −f (r)), rather than r1 . This indeed makes
the analysis a little more complicated. We have
∫
(
)
(
)
X ∗ η + H1′ A−1 −h′′ H ′ + h′2 H ′′ dz1
R
)
(
∫
(
) ∂r 2 A
−1
′ ′
+ X ∗ η + H1′
−
r
∂
r
r 2 (h H ) dz1
2
2A
R
( √ )
= o e− 2D .
For the same reason,
(
)
∫
( + ′ ) ∂z2 A
(
)
∗
−1
X η H1
+ r ∂r z2 H ′ dz1 = o e−D .
2A
R
22
∫
It follows that
R
Similarly,
∫
R
( √ )
(
)
X ∗ η + H1′ E (H1s ) dz1 = o e− 2D .
(
)
( √ )
2
X ∗ η + H1′ 3 (H1 + 1) (H1s + 1) = o e− 2D .
It then follows from Lemma 12 and Lemma 14 that
∫
(
)
X ∗ η + H1′ E (ū)
R
]
[
c0 f ′′
c0 f ′
c0 h′′
√
= − (1 + o (1))
+
3 +
1 + f ′2
r1 1 + f ′2
(1 + f ′2 ) 2
√
( )
− c1 (1 + o (1)) e− 2D(r1 ) + O h′2 + O (h′ f ′′ )
(
)
+ O (h′ f ′′′ ) + O h′ r−1 + O (h′′ f ′′ ) .
This together with Lemma 16 leads to the following equation:
c0 f ′′
√
c0 h′′
c0 f ′
− (1 + o (1)) c1 e− 2D
+ √
(26)
′2
1+f
r1 1 + f ′2
(1 + f ′2 )
( )
(
)
= O (h′′ f ′′ ) + O h′2 + O (h′ f ′′ ) + O h′ r−1 + O (h′ f ′′′ )
(
)
(
)
√
2
+ O ∥ϕ∗ (r1 , ·)∥∞ + O ∥ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O (∥ϕ∗ (r1 , ·)∥∞ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
(
)
(
)
√
)
(
2
+ O ∥∂r1 ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O ∥ϕ∗ (r1 , ·)∥∞ ∂r21 ϕ∗ (r1 , ·)∞
3
2
+
+ O (f ′′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ ) + O (f ′′ ∥∂r1 ϕ∗ (r1 , ·)∥∞ ) + O (f ′′′ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
(
)
(
)
+ O r−1 ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O r−1 f ′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ .
On the other hand,
⊥
∥
Lϕ = [E (ū)] + [E (ū)] + P (ϕ) .
∥
By Lemma 16, [E (ū)] + P (ϕ) is small compared to ϕ. In view of Lemma 17,
⊥
we need to estimate [E (ū)] .
By (19) ,
[
(
)]⊥
⊥
⊥
[E (ū)] = [E (H1 )] + 3 (H1s + 1) H12 − 1
[
]⊥
⊥
2
+ [E (H1s )] + 3 (H1 + 1) (H1s + 1)
.
⊥
Let us analyze the term [E (H1s )] . We know that
(
)
∂z 2 A
E (H1s ) = A−1 (−h′′ H ′ + h′ H ′′ ) +
+ r−1 ∂r z2 H ′
2A
(
)
∂r 2 A
+
− r−1 ∂r r2 h′ H ′ .
2A2
23
Hence in the upper plane, due to the presence of H ′ in each term, we find that
directly that
( √ )
⊥
[E (H1s )] = O e− 2f .
(27)
Note that this is not an optimal estimate and each term in E (H1s ) is mulitplied
by h′ , h′′ , f ′′ or r−1 f ′ . However, these terms are in general not evaluated at r1 .
One also has
[
]⊥
( √ )
2
3 (H1 + 1) (H1s + 1)
= o e− 2D .
The above estimates combined with Lemma 13 and 14 leads to
(
)
(
)
(
)
⊥
⊥
[E (ū)] = [E (H1s )] + O f ′′2 + O r−2 f ′2 + O (h′′ f ′′ ) + O h′ r−1
( √ )
( )
( )
+ O h′′2 + O h′2 + O (h′ f ′′ ) + O (h′ f ′′′ ) + O e− 2D .
(28)
Notice that there is a term f ′′2 appeared in the right hand side. Insert (26)
into (28) we get
(
)
(
)
⊥
⊥
(29)
(E (ū)) = [E (H1s )] + O r−2 f ′2 + O h′ r−1
( √ )
( ′2 )
+ O h + O (h′ f ′′′ ) + O e− 2D
( )
+ O (h′ h′′ ) + O (hf ′′′ ) + O h′′2
(
)
(
)
2
2
+ O ∥ϕ∗ (r1 , ·)∥∞ + O ∥∂r1 ϕ∗ (r1 , ·)∥∞
(
)
2
+ O ∥∂z1 ϕ∗ (r1 , ·)∥∞ .
Note that the term involving h and its derivatives are essentially controlled by
ϕ. Using similar arguments as that of Lemma 17, we find that for s > r1 ,
|ϕ∗ (s, z1 )| ≤ C e−D L∞ ((r1 ,+∞)) + C r−2 f ′2 L∞ ((r1 +∞)) + ∥ϕ∥L∞ (∂Jr ) er1 −s .
1
(30)
s ⊥
)]
appearing
in
(29)
.
Initially,
by
(27),
We remark that there is the
term
[E
(H
1
(
)
it is only of the order O e−f . But we would
( using
) a bootstrap argument to
show that this term is indeed of the order O e−D .
The estimate (30) could be applied repeatedly. This yields that for some
large but fixed constant l,
ϕ (r, z) ≤ Ce−D(r−l) +
≤ Ce−D(r−l) +
[
+C
2
e
C
2
(r − l)
C
+ C ∥ϕ∥L∞ (∂Jr−l ) e−l
2
(r − l)
−D(r−2l)
+
]
1
2
+ ∥ϕ∥L∞ (∂Jr−2l ) e
(r − 2l)
C
≤ Ce−D(r) + 2 + C ∥ϕ∥L∞ (∂Jr ) e−r .
0
r
24
−l
e−l
The last inequality follows from
k
∑
j=1
1
−jl
2e
(r − jl)
C
,
r2
≤
and by Lemma 6, f ′ is small, which together with D = (2 + o (1)) f implies that
k
∑
e−D(r−jl) e−jl ≤ Ce−D(r) .
j=1
This concludes the proof.
Up to now, we have assumed (A) . But a priori, we don’t know whether this
is fulfilled or not. It the sequel, we sketch the proof without this assumption.
The basic principle is, using equation (26) and similar arguments above, one
could prove that assumption (A) indeed holds.
Proof of Proposition 11 without Assumption (A). The main reason that
we √
introduce the Assumption (A) is that in the the error E (ū) is related to
¯
e−2 2d and (we could)control it by e−D . Now without Assumption (A), to handle
√
the term O e−2
2d¯
, we would like to use the following basic geometrical fact
concerning the Fermi coordinate (see [19] for a proof): Using the fact that f ′ (r1 )
is small, there exists a point r̄(r̄ may be equal to r1 ) with
|r̄ − r1 | = o (min {d1 (r1 ) , d2 (r1 )})
such that
min {d1 (r1 ) , d2 (r1 )} ≥
1
.
(r̄)|
|f ′′
(31)
Let us still assume d¯′ ≤ 12 and d¯′′ ≤ C for the moment. Then repeating
the previous arguments, one could show that for r1 > r0 ,
√ √
C
¯
|f ′′ (r1 )| ≤ Ce−r1 +
+ Ce− 2D(r1 ) + C e−2 2d ∞
.
(32)
r1
L (r1 ,+∞)
We claim there exists a universal constant C0 such that
d¯(r1 ) = 3f (r1 ) for r1 > C0 .
(33)
Suppose to the contrary that (33)
is √not satisfied at some√ large point s.
¯
¯
Without loss of generality we assume e−2 2d ∞
= e−2 2d(s) . By (31) ,
L
((s,+∞))
we could find s1 with
|s1 − s| = o (min {d1 (s) , d2 (s)}) = o (f (s))
such that
min {d1 (s) , d2 (s)} >
25
1
.
|f ′′ (s1 )|
(34)
We would like to show that for this point s1 , necessarily
1
≥ 3f (s) .
|f ′′ (s1 )|
(35)
Once this is proved, we then get a contradiction and hence the proof will be
complished.
To prove (35) , we could assume s1 ̸= s, otherwise using estimate (32), we
obtain (35). For the point s1 , we have
√ √
C
¯
+ Ce− 2D(s1 ) + C e−2 2d ∞
s1
L (s1 ,+∞)
√
√
C
¯
+
+ Ce− 2D(s1 ) + Ce−2 2d(s̄1 )
s1
|f ′′ (s1 )| ≤ Ce−s1 +
= Ce−s1
for some point s̄1 ≥ s1 . If Ce−2
√
¯ 1)
2d(s̄
<
|f ′′ (s1 )| ≤ Ce−s1 +
1
2
|f ′′ (s1 )| , then
√
C
+ Ce− 2D(s1 ) ,
s1
which implies (35) . Hence we could assume
Ce−2
√
¯ 1)
2d(s̄
≥
1 ′′
|f (s1 )| .
2
(36)
This in particular implies that
1
d¯(s̄1 ) ≤ √
2 2
(
2C
f ′′ (s1 )
ln
)
≤ ln f (s) .
For the point s̄1 , by (31) , we could find a point s2 with
s2 ≥ s̄1 + o (min {d1 (s̄1 ) , d2 (s̄1 )}) ≥ s1 − ln f (s) ,
such that
min {d1 (s̄1 ) , d2 (s̄1 )} ≥
Note that by (36) ,
f ′′
1
.
(s2 )
√
1 ′′
|f (s1 )| .
2
Now repeat this argument and we get a sequence {sn } with
Ce
2 2
− f ′′
(s )
2
≥
sn+1 ≥ sn − ln
such that
Ce
−
√
2 2
f ′′ (sn+1 )
≥
26
1
,
(s)|
|fn′′
1 ′′
|f (sn )| .
2
(37)
If the process stops at the step n0 < f (s) . Then
f ′′ (sn0 ) ≤ Ce−sn0 +
1
+ e−D(sn0 ) .
sn0
By (37) , sn0 > s − f (s) and D (sn0 ) > f (s) . This then implies (35) . On the
other hand,
if the process doesn’t stop at the step n0 > f (s) , then the sequence
√
ρn := |f ′′2 (s2n )| satisfies
ρn ≥ Ceρn+1 .
(38)
(38) leads to the inequality
ρ1 ≥ Cn20 ≥ Cf 2 (s) ,
This also implies (35) .
Let us we return back to the assumption that d¯′ ≤ 12 and d¯′′ ≤ C. Essentially, this assumption is to make sure that the cutoff function η has bounded
first and second derivatives and thus the error of approximate solution could be
uniformly controlled. However, with the original definition of d¯ this assumption
may not be true. To overcome this difficulty, we should modify the function
¯
d(that
is, one needs to modify the domain Bu ). Let us be more precise. Fix a
small positive constant δ. We define new function dˆ by
{
}
dˆ(r1 ) := inf d¯(s) + δ (r1 − s) : r0 < s < r1 ; d¯(s) − δ (r1 − s) : s > r1 .
Modify dˆ such that it becomes C 2 . We then define the domain Bu using this
ˆ Using similar arguments as before, we could show (33) .
new function d.
Now (33) implies that the size of the Fermi coordinate is actually large
enough for our purpose, that is, Assumption (A) is indeed satisfied.
With all these preparation, we proceed to prove Proposition 4, the main
result of this section.
Proof of Proposition 4 and Theorem 2. We split the proof into several
steps.
Step 1. We first show that there exists a universal constant C, δ > 0 such
that
(√
)
2
p (r) ≥ C +
+ δ ln r.
2
We would like to use equation (26). Note that there is a term involving the
third derivative of f in this equation, although one expects that f ′′′ decays like
r−3 , a priori we don’t have any decay information for it. However, we at least
know that f ′′′ (r1 ) tends to zero as r1 tending to infinity. Our aim is to show
the following estimate(not optimal):
|f ′′′ (r1 )| ≤ Cr1−1 + Ce−D(r1 ) .
(39)
To obtain this estimate, we would like to differentiate equation (26) with respect
to r1 . Using the fact that f (3) and f (4) are small for r1 large(at this stage, we
27
don’t know the decay rate for these two terms, but they are multiplied by terms
related to ϕ), we obtain the following estimate (which is not optimal but enough
for our purpose)
√
f ′′′ + 1 + f ′2 h′′′
( √ )
(
)
= O (h′′ ) + O r1−1 + O e− 2D
(
)
√
+ O (h′ ) + O (∥ϕ∗ (r1 , ·)∥∞ ) + O ∥∂r1 ϕ∗ (r1 , ·)∥∞ e− 2f (r1 )
)
(
+ O ∂r21 ϕ∗ (r1 , ·)∞ + O (∥∂r1 ∂z1 ϕ∗ (r1 , ·)∥∞ ) .
Hence by the C 2,α estimate of ϕ, to prove (39) , it will be suffice to obtain an
estimate for the function h′′′ , which is essentially controlled by ∂r31 ϕ∗ . To achieve
this, we consider the equation
⊥
∥
Lϕ = [E (ū)] + [E (ū)] + P (ϕ) .
(40)
∥
Here we have in mind that by Lemma 16, [E (ū)] is expressed as a small order
⊥
term of ϕ. We also have the estimate (29) for the source term [E (ū)] . Now
∞
differentiate equation (40) with respect to r1 , use the L norm estimate of ϕ
and the fact that
(
)
(
)
⊥
∂r1 [E (ū)] = O r1−1 + O e−D + o (h′′′ ) ,
we find that for r1 large,
3 ∗
∂r ϕ ≤ Cr−1 + Ce−D .
1
1
This then implies (39) .
With the decay estimate of f ′′′ available, equation (26) could be refined to
(
)′
√
(
)
r1 p ′
c1
√
= r1 (1 + o (1)) e− 2D + O r1−2 := I1 + I2 .
(41)
′2
c0
1+p
√
Here p = f + 1 + f ′2 h is the function introduced in Lemma 12. At this stage,
one of the technical difficulty is that this equation involves the function D. As
we mentioned before, we expect that D is very close to 2p. But without a priori
estimate for f ′ and f, |D − 2p| in principle could be large. In any case, we at
least know that D < 2p.
To proceed, we shall fix a small constant δ̄ > 0 which will be determined
later. For any interval (t1 , t2 ) ⊂ (r0 , +∞) , integrating equation (41) in r1 yields
∫ t2
∫ t2
t p′ (t2 )
t1 p′ (t1 )
√ 2
−√
=
I1 +
I2 .
(42)
1 + p′2 (t2 )
1 + p′2 (t1 )
t1
t1
∫t
Keep in mind that t12 I1 is always positive, and by |I2 (r)| ≤ C0 r−2 ,
∫ t2 C0
.
(43)
I2 ≤
t1
t1
28
Set t∗ := max
{ 2C
0
δ̄
}
, r0 . We consider two cases. Case 1:
√
2
√
≥
+ δ̄.
′2
∗
2
1 + p (t )
t∗ p′ (t∗ )
Then by (42) and (43), for all t > t∗ ,
√
Case 2:
√
tp′ (t)
1+
p′2
(t)
≥
2 δ̄
+ .
2
2
√
t∗ p′ (t∗ )
2
√
<
+ δ̄.
′2
∗
2
1 + p (t )
(
)
In this case, let t∗ , t̂ be an interval such that
√
(
)
tp′ (t)
2
√
≤
+ 2δ̄, ∀t ∈ t∗ , t̂ .
′2
2
1 + p (t)
Then the fact that |p′ | = o (1) and a simple integration yield
(√
)
(
)
2
p (t) ≤
+ 3δ̄ ln t + C, ∀t ∈ t∗ , t̂ .
2
This
( ∗ )in particular implies that D (r) − 2f (r) = o (1) and hence for any t1 , t2 ∈
t , t̂ ,
∫ t2
√
C
C
se− 2D(s) ds ≥ √ − √ .
6 2δ̄
6 2δ̄
t1
t1
t2
√
We choose δ̄ such that 6 2δ̄ < 1. Then by inequality (43) , if t2 is large,
∫ t2
∫ t2
I1 +
I2 > 0.
t1
This implies
t1
t p′ (t2 )
t1 p′ (t1 )
√ 2
>√
.
′2
1 + p (t2 )
1 + p′2 (t1 )
Combing the analysis for Case 1 and Case 2, we find that there exist universal
constants δ and C1 such that
√
2
′
rp (r) >
+ δ, r > C1 .
(44)
2
This proves Theorem 2.
With the lower bound (44) available, one could show that
)
(
√
√
e− 2D = e−2 2p + O r−(2+α) .
29
It also follows that
(
)
k
+ O r−1−α ,
r
for some α > 0. Here k is the growth rate of u. Therefore,
(
)
p (r) = k ln r + O r−α + C.
p′ (r) =
With this estimate at hand, we find that un converges to a two-end solution u0
strongly.
3
3.1
Moduli space theory of two-end solutions
Preliminary results
Generally speaking, the structure of the set of bounded entire solutions to the
Allen-Cahn equation could potentially be very complicated. However, if one
impose certain conditions at infinity for the solution, then it could be simpler.
The moduli space theory for multiple-end solutions of the Allen-Cahn equation
in R2 was developed in [9]. This theory tells us that if u is a 2k-end solutions
in R2 and nondegenerate, then around u (in suitable sense), the set of 2k-end
solutions is actually a 2k-dimensional manifold. This fact has been used in an
essential way in the classification of four-end solutions of Allen-Cahn equation
in R2 ([19]). In this section, we would like to develop the corresponding moduli
space theory for two-end solutions in R3 . Our main result states that the moduli
space of two-end solutions has a structure of real analytic variety of formal
dimension 1.
To begin with, let us recall some preliminary results about real analytic operators. Compared to C ∞ operators, real analytic operators has better structures.
We refer to [5], [6], [7] for more details.
Let X and Y be Banach spaces and U an open subset of X. We first recall
the notion of real analytic operator.
Definition 18 A map F : U → Y is real analytic at x0 ∈ U if there exists a
δ > 0 such that
F (x) − F (x0 ) =
+∞
∑
mk ((x − x0 ) , ..., (x − x0 )) , for |x − x0 | < δ,
k=1
where mk is a symmetric k−linear operator, and there exists r > 0 such that
sup rk ∥mk ∥ < +∞.
k>0
The function is said to be real analytic on U if it is real analytic at every point
of U.
30
Let F : U → Y be a real analytic functional. Suppose that dF (x) is a
Fredholm operator of index 1. Assume there exists a map Λ : (0, ε) → Y such
that F (Λ (s)) = 0, and dF (Λ (s)) : X → Y, is surjective for all s ∈ (0, ε) . Let
S = {x ∈ U : F (x) = 0} .
The following theorem has been proved in the book of Buffoni and Toland[5].
Theorem 19 Suppose all bounded closed subsets of S are compact in X. Then
there exists an extension of Λ, denoted by Λ̄ :
(0, +∞) → X,
satisfying: (1) Λ̄ is continuous. (2) The set of points where dF is notsurjective
has no accumulation points. (3) One of the following happens: (i) Λ̄ (s) →
+∞, as s → +∞; (ii) Λ̄ (s) approaches the boundary of U as s tends to infinity;
(iii) Λ̄ ((0, +∞)) is a closed loop.
Basically, this theorem tells us that if one has a real analytic variety which
comes from the zero set of a real analytic operator and its formal dimension is
one, and assume further that on the variety there are some points where the
operator is surjective, then under certain compactness assumption, one could
find a continuous path of solutions where at most finitely many solutions are
degenerate (the linearized operator is not surjective).
3.2
The real analytic structure of the moduli space
Let u be a two-end solution. Then u satisfies (2) and there are constants k and
ck ∈ R such that
∥u (r, ·) − H (· − k ln r − ck )∥L∞ ([0,+∞)) → 0, as r → +∞.
(45)
√
We also assume k > 2. The moduli space theory of noncompact geometric
objects with controlled geometry at infinity(minimal surfaces with finite total
curvature, singular Yamabe metrics, constant mean curvature surfaces with Delaunay ends) has been developed in [21], [23], [25]. In this paper, we will not
investigate the general moduli space theory for the Allen-Cahn equation in dimension three. Instead, we shall only consider those solutions satisfying (2) .
Our aim is to show that the set of solutions of (2) satisfying (45) has the structure of a real analytic variety with formal dimension 1. Additionally, if a solution
u is nondegenerate, then around u, this real analytic variety is actually a one
dimensional real analytic manifold.
Consider the function
(
)
fk,b (r) = k cosh−1 k −1 r + b.
√
where k and b are real parameters, k > 2. Obviously, f represents a vertically translated catenoidal end and we have in mind that f is the asymptotic
31
curve of the nodal line of a solution u. The moduli space theory for two-end
solutions will, roughly speaking, state that around a given solution u there is a
one dimensional family of solutions, with k or b being possible candidates for
the local parameters. To make this statement more precise, we need to analyze
the mapping property of the linearized Allen-Cahn operator around u.
Let us introduce some notations. As in Section 2, we shall also carry out
the analysis through the Fermi coordinate. Let Bu be the domain where the
Fermi coordinate of z = f (r) is well defined(suitably modified if necessary). We
still use η to denote a smooth cutoff function supported in Bu . Similarly, we
have the cutoff function η + supported in Bu ∩ E+ . Similarly, we have the map
X : (r1 , z1 ) → (r, z) .
We need to work in suitable functional spaces S1 , S2 which will be described
now. Let C1 > 0 be a fixed constant and δ > 0 small. A C 2,α function Ξ ∈ S1 ,
if and only if Ξ (r, z) = Ξ (r, −z) and it satisfies the following condition: In E+ ,
when r1 > C1 ,
where
∫
X ∗ Ξ (r1 , z1 ) = p (r1 ) X ∗ η + (r1 , z1 ) H ′ (z1 ) + X ∗ ψ (r1 , z1 ) ,
X ∗ (η + ψ) H ′ dz1 = 0,
2 3 (1 + r1 ) p ∞ + (1 + r1 ) p′ R
L
L∞
4
(1 + r1 ) X ∗ ψ and
4
+ (1 + r1 ) p′′ C 2,α
C 0,α
< +∞,
< +∞.
Similarly, a C 0,α function Ξ ∈ S2 if and only if Ξ (r, z) = Ξ (r, −z) and it
satisfies the following condition: In E+ , when r1 > C1 ,
where
∫
R
X ∗ Ξ (r1 , z1 ) = p (r1 ) X ∗ η + (r1 , z1 ) H ′ (z1 ) + X ∗ ψ (r1 , z1 ) ,
X ∗ (η + ψ) H ′ dz1 = 0,
4 (1 + r1 ) p
C 0,α
and
4
(1 + r1 ) X ∗ ψ < +∞.
C 0,α
< +∞.
Let L be the linearized operator of the Allen-Cahn equation around a twoend solution u, that is,
L = Lu := ∆(r,z) + r−1 ∂r + 1 − 3u2 .
For each point p, let (r1 (p) , z1 (p)) be its Fermi coordinate with respect to
fk,b . Let p′s,t be the point whose Fermi coordinate with respect to the curve
z = fk+s,b+t is still equal to (r1 (p) , z1 (p)) . For p in the upper half plane, now
let Φs,t be the map defined by
(
)
Φs,t (p) = η + p′s,t + 1 − η + p,
32
while for p = (r, z) in the lower half plane, we let
Φs,t (p) = −Φs,t (r, −z) .
In this way, we have defined the family of maps Φs,t which is even with respect to
the r axis. For |s| , |t| small, Φs,t is a diffeomorphism. By definition Φ0,0 (p) = p.
Similarly, we introduce the family of diffeomorphism
Ψs,t (p) := ηp′s,t + (1 − η) p.
Similar arguments as that of Section 4 shows that at far away an axially
symmetric two-end solution u could be written as:
u = H1 + H1s + 1 + ϕ.
Here
H1 = ηH1 + (1 − η)
where
and
H1
,
|H1 |
X ∗ H1 (r1 , z1 ) =H (z1 − h (r1 ))
∫
R
(
)
X ∗ η + ϕ H ′ dz1 = 0.
−4
Notice that ϕ decays like r at infinity.
For (s, t, ψ) ∈ R2 ⊕ S1 , we then define a family of function us,t,ψ such that
at far away
[
]
−1 s
us,t,ψ = H1 ◦Ψ−1
+ (ϕ + ψ) ◦ Φ−1
s,t + H1 ◦Ψs,t
s,t ,
and in a fixed large ball, us,t,ψ = u. Clearly, for u0,0,0 = u. Consider the nonlinear map N : R2 ⊕ S1 → S2 given by
]
[
(s, t, ψ) → ∆us,t,ψ + us,t,ψ − u3s,t,ψ ◦ Φs,t .
The reason that N maps R2 ⊕ S1 into S2 lies in the asymptotic behavior of u,
see Section 2. We have
[
]
[
]
∂s N (s, t, ψ) = Lus,t,ψ ∂s us,t,ψ ◦ Φs,t + ∂s Φs,t · ∇ ∆us,t,ψ + us,t,ψ − u3s,t,ψ ◦ Φs,t ,
[
]
[
]
∂t N (s, t, ψ) = Lus,t,ψ ∂t us,t,ψ ◦ Φs,t + ∂t Φs,t · ∇ ∆us,t,ψ + us,t,ψ − u3s,t,ψ ◦ Φs,t ,
[
]
∂ψ N (s, t, ψ) G = Lus,t,ψ G ◦ Φs,t .
In particular, although DN is not exactly equal to L, it is a small perturbation
of it for s, t small. Observe that when (s, t, ψ) = (0, 0, 0) , DN actually is equal
to L. Let
γ1 = ∂s us,t,ψ |(s,t,ψ)=(0,0,0) ,
γ2 = ∂t us,t,ψ |(s,t,ψ)=(0,0,0) .
33
Introduce the deficiency space
D =span {γ1 , γ2 } .
The next result concerns the mapping property of the linearized operator L and
is the main result of this section.
Proposition 20 The operator L : S1 ⊕ D → S2 is a Fredholm operator of index
1.
Proof. Let us prove the result under the additional assumption that f (0) is
large and ∥f ′ ∥L∞ , ∥f ′′ ∥L∞ are very small. In the general case, we could modify
the function u inside a compact set into a function whose nodal lines are almost
parallel, and use the fact that the corresponding linearized operator is a compact
perturbation(thus the Fredholm index is preserved) of L. We shall split the proof
into several steps.
Step 1. For each function Θ ∈ S2 , we shall find a solution Ξ to the equation
LΞ = Θ.
(46)
By the definition of S2 , in E , for r1 large,
+
X ∗ Θ (r1 , z1 ) = ω (r1 ) X ∗ η + H ′ (z1 ) + X ∗ φ (r1 , z1 ) ,
∫
for some function ω and φ, where R X ∗ (η + φ) H ′ = 0. To solve (46) , adopt the
same notation as in Section 2, it will be suffice to solve the following system
{
∥
∥
(LΞ)1 = Θ1 ,
(47)
⊥
(LΞ) = Θ⊥ .
For convenience, we recall the expression of Lin the Fermi coordinate.
1 ∂z 1 A
1 ∂r 1 A
∂z 1 −
∂r 1
2 A
2 A2
+ r−1 (∂r r1 ∂r1 + ∂r z1 ∂z1 ) + 1 − 3u2 .
L = A−1 ∂r21 + ∂z21 +
Let us first compute the action of L on functions of the form ξ (r1 ) η + H ′ (z1 ) .
Using (12) we get
[
]
(
)
1 ∂r1 A
1
L ξη + H ′ = A−1 ξ ′′ η + H ′ + −
+
ξ′ η+ H ′
2 A2
r (1 + f ′2 ) B
[
]
1 ∂z1 A
f′
+
− √
ξη + H ′′
2 A
r 1 + f ′2
[
(
) ]
+ H ′′′ + 1 − 3u2 H ′ ξη +
+ 2A−1 ξ ′ ∂r1 η + H ′ + A−1 ξ∂r21 η + H ′ + ξ∂z1 η + H ′′ + ξ∂z21 η + H ′
[
]
1 ∂r 1 A
1
+ −
+
ξ∂r1 η + H ′
2 A2
r (1 + f ′2 ) B
[
]
1 ∂z1 A
f′
+
− √
ξ∂z1 η + H ′ .
2 A
r 1 + f ′2
34
The coefficient before ξ ′ has the estimate
−
(
)
1 ∂r 1 A
1
1
+
= + O r−3 .
2 A2
r (1 + f ′2 ) B
r
Next we shall estimate the coefficient before ξ. Recall that
(
)
1 ∂z 1 A
f′
− √
= O r−6 .
′2
2 A
r 1+f
We also have
(
)
(
)
H ′′′ + 1 − 3u2 H ′ = 3 H 2 − u2 H ′
= 3 (H + u) (H − u) H ′ .
Since
)
(
u − H = H (z1 − h (r1 )) − H (z1 ) + O r−2−α
(
)
(
)
= −h (r1 ) H ′ + O r−2−α = O r−2−α ,
(
)
(
)
hence H ′′′ + 1 − 3u2 H ′ = O r−2−α
. As )for those terms involving derivatives
(
+
−2−α
of η , they could be estimated by O r
. Combining all the above estimate,
we get
(
)
(
)
(
)
(
)
1
L ξη + H ′ = A−1 ξ ′′ η + H ′ +
+ O r1−3 ξ ′ η + H ′ + O r−2−α ξ.
r1
Introduce the operator
∫
K1 : ξ →
R
(
)
L ξη + H ′ η + H ′ dz1 .
It will be important to analyze the mapping property of K1 . Let us consider the
equation
K1 ξ = ω.
(48)
This is a second order ODE. Suppose ϑ1 , ϑ2 are two linearly independent solutions of the corresponding homogeneous equation: K1 ϑi = 0. We could assume
ϑ1 is bounded near 0. Let χ be a cutoff function equals 0 in (0, 1) and equals to 1
in (2, +∞) . Then using the variation of parameter formula, one could show that
K1 is an isomorphism from the space S1 ⊕ span {χϑ2 } to S2 . Here a function
p ∈ S1 iff p ∈ C 2,α (R) and
2 3 4
< +∞,
(1 + r1 ) p ∞ + (1 + r1 ) p′ ∞ + max (1 + r1 ) p′′ 0,α
s
L
L
C
(B1 (s))
and a function p ∈ S2 iff p ∈ C 0,α (R) and
4 max (1 + r1 ) p 0,α
s
C
35
(B1 (s))
< +∞.
Define the operator
]s )]⊥
[ (
[
.
K2 : ξ → L ξη + H ′ + ξη + H ′
To solve (47) , it suffices to find solution (ξ, ψ) to the system:
∫
{
K1 ξ = ω − R η + H ′ Lψdz1 ,
⊥
(Lψ) = φ − K2 ξ.
(49)
Let ψ0 be the solution of
⊥
(Lψ0 ) = −K2 (χϑ2 ) ,
∫
satisfying R X ∗ (ψ0 η + ) H ′ dz1 = 0. Note
that
(ln r))
( −2−α
) at infinity, ϑ1 behaves like(O−2−α
or O (1)
.
Hence
K
ϑ
behaves
like
O
r
.
This
implies
that
ψ
=
O
r
.
2
1
0
(
)
∫
Thus R X ∗ (Lψ0 η + ) H ′ dz1 = O r−4 . Thanks to these estimates, one then
could use a fixed point argument to get a pair of solution (ξ, ψ) for (49) with
the form
{
ξ = c1 χϑ2 + ξ0 ,
ψ = c1 ψ0 + ψ̄,
for some constant c1 , where ξ0 ∈ S1 and ψ̄ ∈ S1 .(Note that ψ0 does not belong
to S1 ). Then we get a corresponding solution Ξ to (47) . We emphasize that at
this stage, we still don’t know whether the solution Ξ belongs to S1 , due to the
non-decaying part
[
]s
Ξ̂ := χ (r1 ) ϑ2 (r1 ) η + H ′ (z1 ) + χ (r1 ) ϑ2 (r1 ) η + H ′ (z1 ) + ψ0 .
(50)
Step 2. Investigate the homogeneous equation
Lζ = 0.
(51)
Firstly, we wish to find a nontrivial solution to (51) . To achieve this, we use
the fact that ϑ1 is in the kernel of K1 . A perturbation argument could be applied
similarly as in step 1 to get a function Ξ0 solving LΞ0 = 0,where Ξ0 is around
s
η + ϑ1 (r1 ) H ′ (z1 ) + [η + ϑ1 (r1 ) H ′ (z1 )] . We could also assume Ξ0 satisfying
∫
(
)
(
)
X ∗ Ξ0 η + H ′ (z1 ) dz1 = ϑ1 (r1 ) + δϑ2 (r1 ) + O r1−α ,
R
for certain δ ∈ R and α > 0.
Secondly we show the solution Ξ0 is in some sense unique. For this purpose,
let us assume Ξ′0 is another function solves LΞ′0 = 0 and
∫
(
)
(
)
X ∗ Ξ0 η + H ′ (z1 ) dz1 = ϑ1 (r1 ) + δ ′ ϑ2 (r1 ) + O r1−α ,
R
for certain δ ′ ∈ R and α > 0. Then the function g := Ξ′0 − Ξ0 solves Lg = 0 and
∫
(
)
(
)
X ∗ gη + H ′ (z1 ) dz1 = (δ ′ − δ) ϑ2 (r1 ) + O r1−α .
(52)
R
36
s
We claim that g = 0. Indeed, writing g as η + ξ (r1 ) H ′ (z1 )+[η + ξ (r1 ) H ′ (z1 )] +
φ, where
∫
(
)
X ∗ η + φ H ′ (z1 ) dz1 = 0
R
and φ = O (r−α ) for some α > 0. We have
∫
{
K1 ξ = R η + H ′ Lφdz1 ,
⊥
(Lφ) = −K2 ξ.
Recall that in the Fermi coordinate with respect to fk,b ,
∫
(
)
X ∗ η + H1′ Lφ dz1
R
]
[
∫
(
)
1 ∂r 1 A
1 ∂z1 A
∗
∂z1 −
∂
= − X ∗ η + H1′ A−1 ∂r21 +
r1 φ dz1
2 A
2 A2
R
∫
)
(
− X ∗ η + H1′ r−1 [∂r1 φ∗ ∂r r1 + ∂z1 φ∗ ∂r z1 ] dz1
∫R
[
(
)
(
(
) )]
+
−X ∗ η + H1′ ∂z21 φ∗ + X ∗ η + H1′ 3ū2 − 1 φ dz1 .
R
(
)
Hence R X ∗ (η + H1′ Lφ) dz1 = O r−2−α . By (52) , we could write ξ = βϑ2 (r1 )+
( −α )
ρ, where β ∈ R and ρ (·) = O r1 . By the mapping property of the operator
K1 (note that ϑ1 is in the kernel of K1 ), for some σ ∈ (0, 1) ,
∫
(
)
2+α
α
+ ′
|β| + ∥(1 + r1 ) ρ∥C 2,σ ≤ C 1 + r1
η H Lφdz1 .
(53)
∫
R
C 0,σ
⊥
On the other hand, by the equation (Lφ) = −K2 ξ,
∥
Lφ = −K2 ξ + (Lφ) .
∥
Note that due to the orthogonality condition, (Lφ) is small compared to φ.
Hence by the a priori estimate of the operator L, suitable weighted norm of φ
could be controlled by
o (|β|) + o ∥(1 + r1α ) ρ∥C 2,σ .
This together with (53) yields that g = 0, which implies Ξ′0 = Ξ0 . Hence the
operator L : S1 ⊕ D → S2 has at most one dimensional kernel.
Step 3. To finish the proof, it remains to show that the solutions Ξ in step
1 and Ξ0 in step 2 indeed belong to the space S1 ⊕ D.
Recall that the deficiency space D is spanned by γ1 and γ2 . Consider the
function Lγ1 , i = 1, 2. We know that Lγ2 ∈ S2 . Hence by Step 1, one could find
a solutions gi satisfy
Lgi = Lγi .
37
Note that
gi − ci Ξ̂ ∈ S1
(54)
for some constants ci , where the function Ξ̂ is a non-decaying term introduced
in (50) .
Hence by Step 2 and the asymptotic behavior of gi and γi , gi −γi = di Ξ0 , i =
1, 2, for some constants di . Then by (54) ,
ci Ξ̂ − γi = di Ξ0 , i = 1, 2.
It follows that Ξ̂ − k1,1 γ1 − k1,2 γ2 ∈ S1 and Ξ̄0 − k2,1 γ1 + k2,2 γ2 ∈ S1 , for some
constants ki,j , i, j = 1, 2. The proof is completed.
Having proved the Fredholm property, we proceed to show that the operators
involved are real analytic.
Lemma 21 The map N is real analytic.
Proof. This follows from the fact that ∆ is a linear operator and the function u3 − u is a real analytic function. Note that the subtle point here is that
the definition of N involves the diffeomorphisms Φs,t , Ψs,t . These maps are certainly not real analytic with respect to the r, z variables, since there is a cutoff
function appeared in their definition. However, these diffeomorphisms are indeed real analytic with respect to the parameters s and t, which could be seem
from the explicit expression (11) of the Fermi coordinate(Notice that f depends
analytically on k and b). Indeed, for a point p = (r, z), by definition,
Ψs,t (p) = ηps,t + (1 − η) p.
Recall that the Fermi coordinate of ps,t with respect to fk+s,b+t is equal to
(r1 , z1 ), which is the Fermi coordinate of p with respect to the curve fk+s,b+t .
Hence we have the relations

z1 f ′

r = r1 − √ k,b
,

2
′
1+(fk,b
)
z

 z = fk,b (r1 ) + √ 1 ′ 2 ,
1+(fk,b )
and




 z̄ =
′
z1 fk+s,b+t
√
,
2
′
1+(fk+s,b+t
)
z1
.
fk+s,b+t (r1 ) + √
2
′
1+(fk+s,b+t
)
r̄ = r1 −
The real analyticity follows from these relations.
Definition 22 A solution u is nondegenerate, if and only if the linearized operator L : S1 ⊕ D → S2 is surjective.
By the results of [2] and [12], nondegenerate two-end solutions do exist.
38
Proposition 23 The set of solutions to (2) satisfying (45) has a structure of
real analytic variety of formal dimension 1. Furthermore, if a solution u is
nondegenerate, then locally around u, the solution set is a one dimensional real
analytic manifold.
Proof. The function us,t,ψ is a solution of the Allen-Cahn equation, if and only
if
N (s, t, ψ) = 0.
(55)
Since N (0, 0, 0) = 0, we write equation (55) in the form
∫
1
[DN (ls, lt, lψ) − DN (0, 0, 0)] (s, t, ψ) dl = 0.
DN (0, 0, 0) (s, t, ψ) +
0
Then the result follows from the fact that N is real analytic and of Fredholm
index 1 and the real analytic implicit function theorem(for example, see [5]).
4
Analysis of solutions on the boundary of the
moduli space
Two different types of two-end solutions to equation (2) have been constructed
using Lyapunov-Schmidt reduction method in [2] and [12]. Let us briefly describe these solutions. The first type of solutions is constructed in [2] and has
the property that their nodal curves are close to suitable scaling of a solution to
the Toda system. We
√ call them Toda type
√ solutions. The growth rate of these
solutions is close to 2 (but greater than 2). The second class of solutions are
those constructed in [12]. Their nodal sets are close to the catenoids, which we
know are described by the function εr = cosh (εz) , where ε is a small parameter.
We call them catenoid type solution. The growth rate of these solutions are of
the order ε−1 , hence tends to infinity as ε → 0.
As we discussed in Section 1, we expect that the moduli space of two-end
solutions is diffeomorphic to R. In particular, we expect that there exists a
one-parameter family of solutions, at one end of this family(the “boundary” of
the moduli space), the solutions should be the Toda type solutions, while on
the other end of the moduli space, the solutions should be the catenoid type
solutions.
In this section, we would like to analyze the solutions near the boundary
of the moduli space. Our purpose is to prove that if Pu (recall that Pu is the
intersection point of the nodal set of u with the coordinate
√ axes) is on the z
axis and |Pu | is large, then the growth rate of u is close to 2. (with additional
efforts, one could also show that u is actually a Toda type solution, but the
proof of Theorem 1 don’t need this fact.) We shall also show that if Pu is on
the r axis and |Pu | is large then u is a catenoid type solution.
39
4.1
Analysis of Toda type solutions
We shall first analyze the solutions whose nodal set has two components which
are very far away from each other. We expect that these solutions are Toda
type. Our main result here is
Proposition 24 Let u be two-end solution. Suppose
√ Pu is on the z axis and
|Pu | is large. Then the growth rate of u is close to 2.
We use q (·) = qε (·) to denote the solution of the following Toda equation:
c0 q ′′ +
√
c0 ′
q − c1 e−2 2q = 0, q ′ (0) = 0.
r
(56)
Observe that explicitly,
√
2
qε (r) = q (εr) −
ln ε
2
√
(1+ar2 )
1
1
for some ε > 0, where q (r) = 2√
ln
with a = 2 c2c
. In particular,
8
0
2
√
qε (r)− 2 ln r−Cε tends to 0 as r tends to infinity with Cε a constant depending
on ε. In the sequel, we choose ε such that q (0) = |Pu | .
We shall follow similar notations as that of Section 2. For example, the
nodal curve of u in the upper half plane will be the graph of function f. The
Fermi coordinate with respect to the graph of f will be denoted by (r1 , z1 ) . We
also have the cutoff functions η, η + , and the solution u will be written the form
u = ū + ϕ, where ū is an approximate solution:
2
ū = H1 + H1s + 1,
with the function H1 defined similarly as that of Section 2 using the cutoff
function η and the heteroclinic solution H.
The main idea of the proof is to compare f with the solution qε of (56) ,
by analyzing the equation satisfied by f. The main step will be establishing
suitable decay estimate for the function ϕ, as we have already done in Section 2.
Our starting point is the fact that if Pu is on the z axis and |Pu | is large, then
∥f ′ ∥L∞ ((0,+∞)) will be small. This follows from an application of the balancing
formula.
We shall get a L∞ estimate for ϕ. In the following, ∥·∥∞ stands for the
∞
L ((0 + ∞)) norm. To simplify the notations, we only consider the case that
the radius of the Fermi coordinate is large enough.
Lemma 25 Suppose u satisfies the assumption of Proposition 24. Then
′2 √
f −2 2f (0)
∥ϕ∥∞ ≤ C .
r2 + Ce
∞
40
Proof. We only sketch the proof, since many computations will be similar to
that of Section 2.
Recall that ϕ satisfies the equation
∥
⊥
Lϕ = [E (ū)] + [E (ū)] + P (ϕ) .
∥
∥
(57)
∥
P (ϕ) is a higher order term of ϕ, and [E (ū)] = E (ū)1 + E (ū)2 is also small
compared to ϕ. As in Section 2,
)
(
)
)
(
(
⊥
⊥
[E (ū)] = [E (H1s )] + O f ′′2 + O r−2 f ′2 + O (h′′ f ′′ ) + O h′ r−1
(58)
( √ )
( ′2 )
( ′′2 )
+ O h + O (h′ f ′′ ) + O (h′ f ′′′ ) + O e− 2D .
+O h
Project equation (57) onto η + H1′ , we could show
c0 f ′′
√
c0 h′′
c0 f ′
− 2D
√
−
(1
+
o
(1))
c
e
+
(59)
1
1 + f ′2
r1 1 + f ′2
(1 + f ′2 )
( )
(
)
= O (h′′ f ′′ ) + O h′2 + O (h′ f ′′ ) + O h′ r−1 + O (h′ f ′′′ )
)
(
)
(
√
2
+ O ∥ϕ∗ (r1 , ·)∥∞ + O ∥ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O (∥ϕ∗ (r1 , ·)∥∞ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
(
)
(
)
√
)
(
2
+ O ∥∂r1 ϕ∗ (r1 , ·)∥∞ e− 2f (r1 ) + O ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O ∥ϕ∗ (r1 , ·)∥∞ ∂r21 ϕ∗ (r1 , ·)∞
3
2
+
+ O (f ′′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ ) + O (f ′′ ∥∂r1 ϕ∗ (r1 , ·)∥∞ ) + O (f ′′′ ∥∂r1 ϕ∗ (r1 , ·)∥∞ )
(
)
(
)
+ O r−1 ∥∂r1 ϕ∗ (r1 , ·)∥∞ + O r−1 f ′ ∥∂z1 ϕ∗ (r1 , ·)∥∞ .
Insert this into (58) , we obtain
(
)
(
)
⊥
⊥
(E (ū)) = [E (H1s )] + O r−2 f ′2 + O h′ r−1
( √ )
( )
+ O h′2 + O (h′ f ′′′ ) + O e− 2D
( )
+ O (h′ h′′ ) + O (hf ′′′ ) + O h′′2
(
)
(
)
2
2
+ O ∥ϕ∗ (r1 , ·)∥∞ + O ∥∂r1 ϕ∗ (r1 , ·)∥∞
(
)
2
+ O ∥∂z1 ϕ∗ (r1 , ·)∥∞ .
Then the a priori estimate of the operator L tells us that
′2 √
f −2 2f (0)
∥ϕ∥∞ ≤ C .
r2 + Ce
∞
Our next aim is to estimate the L∞ norm of
f ′ (r)
r .
Lemma 26 Let ε be introduced above, then
′
f + ∥f ′′ ∥ ≤ Cε2 .
∞
r
∞
41
Proof. The starting point of the proof is still equation (59) . Applying Lemma
25, we infer from (59) that
c0 f ′′
3
2
(1 + f ′2 )
( )
= O ε2 .
Let t0 be a point where
c0 f ′
+
r1 (1 + f ′2 )
1
2
− c1 e−
√
2D
(60)
′
f ′ (t0 ) f .
=
t0
r ∞
′
This point exists because fr → 0, as r → +∞.
At the point t0 , by (60) , f will satisfy
c0 f ′′ (t0 ) + c0
( )
= O ε2 .
We claim
√
f ′ (t0 )
− c1 e− 2D(t0 )
t0
f ′′ (t0 ) ≥ 0.
(61)
(62)
′′
Indeed, if this is not true, then due to the fact that f (0) ≥ 0, there will be
another point t1 < t0 such that f ′′ (t1 ) = 0, and f ′′ (t) < 0, t ∈ (t1 , t0 ) . Since
f ′ ≥ 0, we find
( ′
)′
f (r)
f ′′ − r−1 f ′
=
< 0, r ∈ (t1 , t0 ) .
r
r
′
′
This contradicts with the fact that f t(t00 ) = fr . From (61) and (62), it may
∞
be concluded that
( )
c1 √
f ′ (t0 )
= e− 2D(t0 ) − f ′′ (t0 ) + O ε2
t0
c0
√
≤ Ce−2
2f (0)
≤ Cε2 .
(63)
Here we have used the fact that D (t0 ) ≥ 2f (0) . This proves estimate for the
′
L∞ norm of fr . The estimate for f ′′ is a direct consequence of (63) and (61) .
√
Let p = f + 1 + f ′2 h. Next we show that p is indeed
to the solution
close
q of the Toda equation in a large interval. Set b = bε = lnεε .
Lemma 27 There exists α > 0, such that
|p (r) − q (r)| ≤ Cεα , r ∈ (0, b) .
( )
(
) √
Proof. Denote by e the function p ε−1 r + 22 ln ε − q (r) . Then e (0) = O ε2
and e′ (0) = 0. By√the previous lemmas, ∥ϕ∥∞ ≤ ε2 . Using this fact, we see that
the function p = 1 + f ′2 h + f satisfies the equation
c0 p′′
(1 + p′2 )
3
2
+
c0 p′
r1 (1 + p′2 )
1
2
42
− c1 e−
√
2D
(
)
= O ε2+α .
In particular, this combined with the fact that D − 2f = O (εα ) yields
p′′ +
√
(
)
p′
− e−2 2p = O ε2+α .
r1
From this equation, we infer that in the region where e is o (1) , e satisfies
e′′ +
√
( )
e′
− e−2 2q e = O e2 + O (εα ) .
r
The conclusion of the lemma then follows from the variation of parameters
formula.
√
By definition q (r) = q (εr) − 22 ln ε, hence rq ′ (r) = εrq′ (εr) , which implies
that
√
bq ′ (b) = 2 + o (1) .
Notice that in Lemma 27, we actually could also estimate |e′ | ≤ Cεα . From this,
we infer that
√
bf ′ (b) = 2 + o (1) .
(64)
Now we are in a position to prove the main result of this section.
Proof of Proposition 24. Similar arguments as before yields
|ϕ (r, z)| ≤ C
1
+ Ce−D(r) .
1 + r2
Equation (59) then becomes
p′′
(1 +
p′2 )
3
2
+
p′
r1 (1 +
p′2 )
1
2
= (1 + o (1)) e−
√
2D(r1 )
(
)
+ O r1−3 .
Integrating from t0 to t1 leads to
(
t1 p′ (t1 )
2
1 + p′ (t1 )
) 12 − (
t0 p′ (t0 )
2
1 + p′ (t0 )
∫
) 12 = (1 + o (1))
t1
e−
√
2D(s)
( )
sds + O t−1
.
0
t0
This together with (64) tells us that
√
p′ (r) r ≥ 2 + o (1) for r > b.
Let δ > 0 be a fixed small constant. We claim that when ε is small,
√
p′ (r) r ≤ 2 + δ for r > b.
(65)
√
Indeed, suppose (b, b∗ ) is the maximal interval where p′ (r) r ≤ 2 + δ. Then in
this interval, elementary geometrical facts implies that
(√
) ∫ r ds
√
2 + o (1)
−C
D (r) ≥ 2 |ln ε| + 2q (b) + 2
s
b
(
)
√
√
= 2 |ln ε| + 2q (b) + 2
2 + o (1) (ln r − ln b) − C.
43
Therefore we have
(
b∗ p′ (b∗ )
2
1 + p′ (b∗ )
) 12 − (
∫
bp′ (b)
1 + p′ (b)
2
) 12
b∗
= (1 + o (1))
e−
√
2D(s)
)
(
sds + O b−1
b
(
)
= O b−1 .
This implies that when
ε is small, b∗ = +∞. Applying (65) , we finally get
√
′
limr→+∞ p (r) r = 2 + o (1) . The proof is thus completed.
4.2
Uniqueness of catenoid type solutions
Beside the planes, catenoid is the first example of embedded minimal surfaces
with finite total curvature; it is rotationally symmetric with respect to its axis
and the only minimal surface of revolution(up to a homothety). In the (r, z)
coordinate, the one-parameter family of catenoids Cε could be represented by
the function
εr = cosh (εz) ,
with ε > 0 being the parameter. As we mentioned before, in [12], for each ε
sufficiently small, a solution uε of the Allen-Cahn equation is constructed. The
nodal set of uε is close to
( the
) catenoid Cε . In particular, Puε is on the r axis,
|P(uε | is) of the order O ε−1 , and the growth rate of uε is also of the order
O ε−1 .
In this section, we wish to prove that this one-parameter family of solutions
uε is unique. This is the content of the following
Proposition 28 Let u be a two-end solution of the Allen-Cahn equation. Suppose Pu is on the r axis and |Pu | is large. Then there exists a small ε′ > 0 such
that u = uε′ .
The nodal curve Nu in E+ could be written as
{(r, z) : z = f (r)} ,
where f is a function
f : [|Pu | , +∞) → R.
Note that by the results in the previous section, f is asymptotic to c1 ln r + c2
as r → +∞ for some constants c1 and c2 . One could also write Nu ∩ E+ =
{(r, z) : r = g (z)} for some function g : [0, +∞) → R.
−1
For convenience we introduce the parameter ε = |Pu | . Observe that by
the assumption of Proposition 28, ε is small, thus by the validity of De Giorgi
conjecture in R3 , locally around the nodal curve, u looks like the heteroclinic
solution.
Let l be a large but fixed constant. As a preliminary step, we would like to
get some rough information about the slope of the function f.
44
Lemma 29 We have
f ′ (r) >
[
]
C
+ o (1) , r ∈ ε−1 , lε−1 .
l
Proof. Let X = (0, 0, 1) be a constant vector field. We have
∫
lε−1
0
On the other hand,
∫
{[
∂Ω∩{z>0}
( −1 )
′
∼f
lε
[
]
1 2
ur + F (u) rdr ≥ Clε−1 .
2
]
}
1 2
ur + F (u) X · v − (∇u · X) (∇u · v) dS
2
lε−1 + o (1) lε−1 .
Combine these two estimates and use the balancing formula, we get the desired
result.
To get more precise information, we again need to work in the Fermi coordinate (s, t) around the nodal curve Nu , where s is the signed distance to Nu
and t is a parametrization of Nu .
We slightly abuse the notation and still use Bu to denote the maximal domain
where the Fermi coordinate is well-defined. Similarly as before, let η be a
cutoff function supported in Bu , with ∇η supported near the boundary of Bu .
Introduce the approximate solution
ū (r, z) = ηH + (1 − η)
H
,
|H|
where H is defined through X ∗ H (s, t) = H (s − h (t)) . Here X is the map
(s, t) → (r, z) . Write u = ū+ϕ. The small function h is chosen such that ϕ
satisfies the orthogonal condition
∫
X ∗ (ηϕH′ ) ds = 0,
R
∗
′
′
where X H (s, t) = H (s − h (t)) .
the solution u, we firstly study Nu in the region where r ∈
( −1To analyze
)
ε , lε−1 . In this region, many calculations will be explicit to in the Fermi
coordinate (x, y) with respect to the graph of the function g, which is

 r = g (y) + √ x ′2 ,
 z = y − √ xg
′
1+g
1+g ′2
.
Hence x is the signed distance and y is a parametrization of the curve.
We shall estimate the L∞ norm of the perturbation term ϕ.
45
Lemma 30 Suppose u satisfies the assumption of Proposition 28 and ϕ is defined above. Then
∥ϕ∥∞ ≤ Cε2 .
Proof. We only consider the case that the size of the Fermi coordinate is large
enough. The general case could be handled using the arguments of Section 2.
Again many computations here are similar as before.
(
)
We need to analyze E (ū) . First of all, consider the case r ∈ ε−1 , lε−1 . In
the Fermi coordinate, the error of the approximate solution has the form
(
)
H ′′ h′2 − H ′ h′′
∂x A ∂r x
E (ū) =
+
+
H′
A
2A
r
(
)
∂y A ∂r y
+
−
H ′ h′ .
2A2
r
Keep in mind that H ′ is evaluated at x − h (y) , not x. By Lemma 29, |g ′ | ≤ C,
hence
√ 1 ′2
∂r x
1+g
=
r
g+ √ x
1+g ′2
(
)
1
x
= √
+ O g −3 .
− 2
′2
g (1 + g )
g 1 + g ′2
From |g ′′ | = o (1), we infer that
( )
∂x A
g ′′
(g ′′ ) x
=−
+ O g ′′3 .
3 +
3
2A
(1 + g ′2 )
(1 + g ′2 ) 2
2
It follows from these expansions that the projection of E (ū) onto H′ has the
form
∫
c0 g ′′
c
√ 0
X ∗ (ηH′ E (ū)) dx = −
3 +
′2
2
g
1
+ g ′2
R
(1 + g )
∫
′2
( )
H
− h′′
+ O h′2 + O (h′ g ′′ )
R A
( )
(
)
+ O (h′ g ′′′ ) + O g ′′3 + O g −3
(
)
(
)
+ O hg ′′2 + O hg −2 .
46
Set h̃ (y) =
−
√
1 + g ′2 h (y) and p1 (y) = g (y) + h̃ (y) . Then
g ′′
1
h′′
√
−
+
1 + g ′2
g
1 + g ′2
h̃′′
g ′′
1
2h̃′
=−
+ −
−√
′2
′2
1+g
g 1+g
1 + g ′2
(
)′′
h̃
1
√
−√
′2
1+g
1 + g ′2
(
1
)′
√
1 + g ′2
1
p′′1
+
+ O (g ′′ h) + O (g ′′ h′ ) + O (g ′′′ h) + O (g ′′′ h′ )
1 + p′2
p
1
1
( )
(
)
+ O (g ′′ h) + O (g ′′ h′ ) + O h2 + O (hh′ ) + O g −1 h .
=−
It follows that
∫
√
c0
c0 p′′1
′2
1+g
X ∗ (ηH′ E (ū)) dx = −
′2 + p
1
+
p
1
R
1
+ O (g ′′ h) + O (g ′′ h′ ) + O (g ′′′ h) + O (g ′′′ h′ )
( )
(
)
+ O (g ′′ h) + O (g ′′ h′ ) + O h2 + O (hh′ ) + O g −1 h
( )
( )
(
)
+ O h′2 + O g ′′3 + O g −3 + O (h′ h′′ ) .
On the other hand, we could also show that
∫
)
(
) (
X ∗ (ηH′ E (ū)) dx = o (∥ϕ∗ (y, ·)∥C 2 )+o ∥∂y ϕ∗ (y, ·)∥C 0 +o ∂y2 ϕ∗ (y, ·)C 0 .
R
(66)
Here the norm is taken as a function of x variable. As a consequence of the
above two equations, we have
−
( )
( )
1
p′′1
+
= O h′2 + O (h′ g ′′ ) + O (h′ g ′′′ ) + O g ′′3
′2
1 + p1
p1
(
)
(
)
(
)
+ O g −3 + O hg ′′2 + O hg −2 + O (h′ h′′ )
(
)
(
)
+ o (∥ϕ∗ (y, ·)∥C 2 ) + o ∥∂y ϕ∗ (y, ·)∥C 0 + o ∂y2 ϕ∗ (y, ·)C 0 .
(67)
Next we consider the case r > ε−1 l. In this case, one could use the Fermi
coordinate (r1 , z1 ) with respect the curve z = f (r) . The analysis of E (ū) in
this case is almost same as that of the previous sections and we omit the details.
Now we write the equation satisfied by ϕ into the form
⊥
∥
Lϕ = [E (ū)] + [E (ū)] + P (ϕ) ,
where
∫
∥
[E (ū)] =
R
X ∗ (ηH′ E (ū)) dx
⊥
∥
∫
ηH′ and [E (ū)] = E (ū) − [E (ū)] .
2 H′2 dx
η
R
47
In terms of Fermi coordinate, in the region where |g ′ | ≤ C,
(
)
(
)
2
⊥
[E (ū)] = O |p′′1 | + O p−2
.
1
)
(
⊥
In the region where |f ′ | ≤ C, [E (ū)] = O r1−2 . By the a priori estimate of
L, we could obtain
⊥
∥ϕ∥∞ ≤ [E (ū)] ≤ Cε2 .
∞
⊥
We emphasize here that [E (ū)] should be estimated in the whole plane. It
∞
(
)
2
⊥
is also worth mentioning that the term O |p′′1 | in [E (ū)] should be handled
using equation (67) .
(
)
Our next aim is to show that in the interval ε−1 , lε−1 , the function r =
p1 (z) is close to the function r = ε−1 cosh (εz) . This is the content of the
following
(
)
Lemma 31 For z ∈ 0, f (l) ε−1 ,
p1 (z) − ε−1 cosh (εz) ≤ Cε.
Proof. From equation (67) and the estimate of ϕ, we deduce that the function
p1 satisfies
( )
1
p′′1
−
= O ε3 .
′2
1 + p1
p1
(
)
At this stage, we introduce the scaled function p̄1 (z) = εp1 ε−1 z . Then
(
)
p̄′1 (z) = p′1 ε−1 z ,
(
)
p̄′′1 (z) = ε−1 p′′1 ε−1 z .
It follows that
p̄′′1 −
( )
1 + p̄′2
1
= O ε2 , z ∈ (0, l) .
p̄1
(68)
Observe that the function cosh z satisfies the equation
′′
(cosh z) −
′2
1 + (cosh z)
cosh z
= 0.
(69)
Let ω (z) = p̄1 (z) − cosh z. Then
ω (0) = p̄1 (0) − 1 = εp1 (0) − 1
(
)
( )
= ε ε−1 + h (0) − 1 = O ε2 .
We claim that |ω (z)| ≤ Cε2 for z ∈ (0, l) . Indeed, subtracting equation (68)
with (69) , we get
( )
( )
( )
ω ′′ − 2 tanh zω ′ + ω = O ε2 + O ω 2 + O ω ′2 := ψ (z) .
48
Let ξ1 and ξ2 be two linearly independent solutions of the homogeneous equation:
ξi′′ − 2 tanh zξi′ + ξi = 0, i = 1, 2.
Explicitly, we can choose
ξ1 (z) = cosh′ z = sinh z,
(
)
ξ2 (z) = ∂ε ε−1 cosh εz |ε=1 = − cosh z + z sinh z.
The Wronsky of these two solutions are
sinh z
W (z) := − cosh z + z sinh z
cosh z = cosh2 z.
z cosh z By the variation of parameters formula,
∫ z
∫ z
( )
ξ1 (s) ψ (s)
ξ2 (s) ψ (s)
ω (z) = ξ2 (z)
ds − ξ1 (z)
ds + O ε2 .
W (s)
W (s)
0
0
The desired estimate comes from this formula.
√
Proof of Proposition 28. Let p2 = f + 1 + f ′2 h. For r > lε−1 , projecting
E (ū) on ηH′ and perform similar calculation as in Section 2, we could estimate
the perturbation ϕ in algebraically weighted norm(Remember that h could be
controlled by ϕ). This leads to the equation:
(
r1 p′2
)′
√
1 + p′2
2
(
=O
ε2
1 + ε2 r 2
)
.
Introduce the scaling of p2 :
(
)
p̄2 (r1 ) = εp2 ε−1 r1 .
We find that in (l, +∞) , p̄2 satisfies has the equation
(
√
r1 p̄′2
1 + p̄′2
2
)′
(
=O
ε2
1 + r2
)
.
From this equation, we deduce
( )
lim r1 p̄′2 (r1 ) − lp̄′2 (l) = O ε2 .
r1 →+∞
On the other hand, by Lemma 31,
( )
lp̄′2 (l) − 1 = O ε2 .
Hence
( )
lim r1 p̄′2 (r1 ) = 1 + O ε2 .
r1 →+∞
49
(70)
Consequently, the growth rate of u is equal to
lim r1 p′2 (r1 ) =
1
1
+ O (ε) := ′ .
ε
ε
( )
Obviously, ε′ = ε + O ε3 .
Now let r = ε′−1 cosh (ε′ z) be the catenoid with the same growth rate as
u. Then there is a solution uε′ with the slope ε′−1 whose nodal line is close
to this catenoid(the nodal line in E+ decaying algebraically to a vertical small
translation of the catenoid. Actually, the error could be estimated by some
positive power of ε. Our aim is to show that u = uε′ . To see this, we need to
estimate the function p1 (z) − ε′−1 cosh (ε′ z) and p2 (r1 ) − ε′−1 cosh−1 (ε′ r1 ) .
Note that
(
)
(
( ))
(
( ) )
εε′−1 cosh ε′ ε−1 z − cosh z = 1 + O ε2 cosh 1 + O ε2 z − cosh z
( )
= O ε2 .
(
)
( )
Hence using Lemma 31, we obtain p̄1 (z) − εε′−1 cosh ε′ ε−1 z = O ε2 . This
implies that
p1 (z) − ε′−1 cosh (ε′ z) = O (ε) .
Using (70) , we then find that p2 (r1 ) − ε′−1 cosh−1 (ε′ r1 ) = O (ε) .
Therefore the nodal line of u is close to the nodal line of uε′ at the order O (ε) .
Then, applying the mapping property of the Jacobi operator of the catenoid, a
contraction mapping argument shows that u = uε′ . We refer to [19] for similar
arguments in the case of four-end solutions in R2 .
5
Concluding the Proof of Theorem 1
In this section, combining the results of the previous sections, we would like to
finish the proof of Theorem 1.
Let M be the set of √
all two-end solutions to the Allen-Cahn equation with
growth rate larger than 2. Consider a catenoid type solution uε arising from a
largely dilated catenoid. As we mentioned in Section 3, uε is nondegenerate.By
Proposition 23, locally around uε , the set of two-end solutions is a one dimensional real analytic manifold, which we denote as the image of a map ϱ
ϱ : (−δ, δ) → M,
for some small δ > 0. Using the compactness result in Section 2 and by the
structure theorem (19), ϱ has a global continuation:
ϱ : (−δ, +∞) → M.
√
We claim that the growth rate of the solution ϱ (t) tends to 2 as t → +∞.
Indeed, as t → +∞, Pϱ(t) could not remain bounded. (Recall that Pϱ(t) is
the intersection of the nodal set of ϱ (t) will the r or z axis.) Otherwise by
50
the compactness, the image of ϱ will be a closed loop, which could not be
true since the family of catenoid type solutions uε are not compact. Also, as
t → +∞, Pϱ(t) could not be on the r axis, this follows from the uniqueness
of catenoid
type
solutions, Proposition 28. Therefore, Pϱ(t) will be on the z
axis and Pϱ(t) → +∞, as t → +∞. By the analysis of Toda type solutions,
√
Proposition 24, the growth rate of the solution ϱ (t) will go to 2 as t → +∞.
This finishes the proof.
6
6.1
Appendix
Monotonicity of two-end solutions
In this appendix, we sketch the proof of monotonicity of two-end solutions(Proposition
3 in Section 2) using the moving plane argument. Essentially, we follow the proof
of monotonicity for four-end solutions of 2D Allen-Cahn equation in [14]. But
their is a slight difference here. Namely for the two-end solutions in dimension three, to start the moving procedure at infinity, one need to have suitable
control of the asymptotic behavior of the solution(estimate (73) below), while
this is not needed in dimension two case. The asymptotic expansion we need is
provided by the results of Section 2.
Proof of Proposition 3. Let u be a two-end solution. We first prove its monotonicity in the r direction. To use the moving plane machinery,
we will work
(√
) in
2
2
the usual Euclidean coordinate (x, y, z) . Set U (x, y, z) := u
x + y ,z =
u (r, z) .
√
Suppose the growth rate of u is equal to k > 2. Hence the asymptotic
curve of its nodal line in the first quadrant is
z = k ln r + b
for some b ∈ R.
Let x0 ≥ 0 be a parameter and define
Ū (x, y, z) = U (x, y, z; x0 ) := U (2x0 − x, y, z) .
Certainly Ū (x0 , y, z) = U (x0 , y, z) . Note that Ū actually depends on the parameter x0 .
The first step is to show that the moving plane procedure could be started
at +∞. We claim that for x0 large enough, say x0 > a0 ,
Ū (x, y, z; x0 ) < U (x, y, z) , for x < x0 .
First of all, we consider the region where |Ū | is not close to 1. Recall that the
nodal set of Ū is close to
√
2
z = k ln (2x0 − x) + y 2 + b.
51
We have
√
√
2
(2x0 − x) + y 2 − ln x2 + y 2
(
)
1
4x20 − 4x0 x
= ln 1 +
.
2
r2
ln
Note that for x < x0 ,
If
4x20 −4x0 x
r2
4x20 −4x0 x
r2
(71)
> 0. Fix a small positive constant ϵ.
> ϵ, then by (71) ,
√
√
1
2
ln (2x0 − x) + y 2 − ln x2 + y 2 > ln (1 + ϵ) .
2
By the results of Section 2, for r large, we have
(
)
u (r, z) − H (z1 ) = O r−2 .
(72)
(73)
Here z1 is the signed distance of (r, z) to the nodal line. As a consequence
Ū (x, y, z; x0 ) − U (x, y, z)
(√
)
2
=u
(2x0 − x) + y 2 , z − u (r, z)
([
]
)
√
2
=H
z − k ln (2x0 − x) + y 2 − b cos θ1
([
]
)
√
(
)
− H z − k ln x2 + y 2 − b cos θ2 + O r−2 ,
(
)
where θi = O r−1 . It follows from (72) that
(
)
√
2
2
Ū (x, y, z; x0 ) − U (x, y, z) ≤ H z − k ln (2x0 − x) + y − b
(
)
√
(
)
− H z − k ln x2 + y 2 − b + O r−2
< 0,
for r large enough.
4x2 −4x x
If 0 r2 0 ∈ (0, ε) , then
(
)
√
2
2
2
√
2x
−
2x
x
x
(x
−
x
)
0
0
0
2
0
0
ln (2x0 − x) + y 2 − ln x2 + y 2 =
+O
. (74)
r2
r4
There are two possible cases. Case 1: x ∈ (−∞, x0 − 1) . In this case, by (74) ,
√
√
x0
2
ln (2x0 − x) + y 2 − ln x2 + y 2 ≥ 2 .
r
This together with (73) implies that for x0 large, Ū < U. Case 2: x ∈ (x0 − 1, x0 ) .
In this case, by the estimate
(
)
∂r (u (r, z) − H (z1 )) = O r−2 ,
52
we find that(here one should also use some estimates of the Fermi coordinate)
Ū − U ≤ −C
(
)
x20 − x0 x
+ O r−2 (x0 − x)
r2
for certain positive constant C. Hence if we choose x0 large, Ū <
U. Therefore,
to prove the claim, it remains to consider the region where Ū ∼ 1. Observe
that for x0 large enough, in the region where Ū ∼ 1, U is also close to 1. Now
let φ := Ū − U. Then in the region x < x0 , φ satisfies
(
)
−∆(x,y,z) φ + Ū 2 + Ū U + U 2 − 1 φ = 0.
Since lim sup|(x,y,z)|→+∞ φ ≤ 0, by the maximum principle, φ (x, y, z) < 0, for
x < x0 . This proves the claim.
In the second step, we define
{
}
x∗ = inf x̄ : Ū (x, y, z; x0 ) < U (x, y, z) for x < x0 and x0 ∈ (x̄, a0 ) .
We show that x∗ = 0. To see this, we first prove that r∗ < a0 . Indeed, by the
first step,
Ū (x, y, z; a0 ) < U (x, y, z) for x < a0 ,
Ū (a0 , y, z; a0 ) = U (a0 , y, z) .
(
)
Hence by the Hopf Lemma, ∂x Ū (·; a0 ) − U > 0 for x = a0 . Then standard arguments together with the asymptotic behavior of U implies that Ū (x, y, z; x0 ) <
U, for x < x0 and x0 sufficiently close to a0 . Then one could use this type of
arguments to show that x∗ = 0. (The plane could be moved to the left until the
inequality Ū < U is violated at infinity.)
The monotonicity in z direction could be proved similarly using moving
plane. This completes the proof.
Acknowledgement Y. Liu is partially supported by NSFC grant 11101141
and the Fundamental Research Funds for the Central Universities 13MS39. J.
Wei is partially supported by NSERC of Canada.
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